 OK? So let me recall the fundamental solution for the Laplace equation. So we found that maybe the notation was phi from the fundamental solution. Phi of x was equal to constant. The log, this was the notation, yes. And the log of modulus of x. And then another constant here. This is n equal to 2. There is no minus. Now we have to look. OK, so there is a minus. So it was OK. And then 1 over n minus 2 omega n, 1 to the x n minus 2. n bigger or equal to n 3. Actually, we can also, so these are singular functions. Maybe we can also see in one dimension. So these are radial functions, which solve the Laplace equation out of the origin, OK? So these are harmonic functions out of the origin, radial. OK, and then we can also take this. Maybe we can add this in one dimension. So let us see in one dimension what is the function phi. So phi is, so minus phi of x in one dimension is just equal to x over 2. Therefore, it's something like this, something like this. Therefore, the derivative of minus phi prime, so this is x. So the derivative here is negative minus 1 half, 1 half, which is, of course, an L1 log is locally integrable. However, phi second is, so minus phi second is not yet defined as far as we know. So phi second would be something like 0, then rapidly increasing here and 0 again. So I don't want to say what is phi second minus phi second for the moment. I don't want to say what it is. Actually, it's called the Dirac delta. We will see what does it mean. In any case, you see now the jump here is 1, the length of the jump. And this is the reason for this constant 1 half in front, essentially, so minus phi sort of convex function like this, minus phi. And the second derivative of minus phi is somehow concentrated at this point with the sign. I mean, this is a convex function. Now the log of modulus of x, so the log of modulus of x with the minus, well, this, again, is a singular function of this four near the origin of this four. We have, and also this, the integrability, we have already checked that the integrability property of this in two dimensions. So this is L1 log. This singularity can be integrated. Also, this can be integrated. This already we already know. Maybe this was an exercise. The gradient also can be integrated locally around the origin, but the second derivative cannot. This may be where exercise is. Now in two dimensions, some remarks. So if remark, if f is holomorphic in omega, then nf, let us write it as u plus ev, then umv are harmonic. This may be a homework if you want. Conversely, assume that u is harmonic in omega. We look for function f of the form u plus iv such that f is holomorphic. Let us see whether this is possible or not. So we look for v such that this system holds. Now let us consider the following differential form. Consider the following one form. So given u, we can consider this one form here. So compute the differential of this one form, which is minus u, y, x, dx wedge dx, which is 0, minus u, y, y, dy, wedge dx, plus u, x, x, dx wedge dy, plus u, x, y, dy wedge dy. And so this is 0. This is 0. And we have that this is equal to u, x, x, plus u, y, y, dx, dy, which is 0 because u is harmonic. So if u is harmonic in omega, then this differential form is closed. So omega is closed in omega. Now, if we assume that omega is simply connected, so conversely, assume that u is harmonic in omega. And so let us add also the following assumption to continue on the argument here. Assume, suppose, omega simply connected. So we have that omega is closed in omega, capital omega. But capital omega at the moment now is we have supposed it to be simply connected. And therefore, omega is exact. Since capital omega is simply connected, omega is exact. Hence, it admits a potential. Therefore, there exists, say, v, function v, smooth, such that omega is equal to dv. What does it mean, omega equal dv? This means vx dx plus dy dy. But omega was minus uy dx plus ux dy. And so this must be equal to this. Hence, we find that minus uy must be equal to vx. And the coefficients ux must be equal to vy. Therefore, we have solved the Cauchy-Riemann equations. So v is the function that we are looking for. It's exactly what we are looking for. With this v, the Cauchy-Riemann equations are satisfied. So we conclude that, no, no, v is not unique. It's determined up to a real constant. So v is determined up a real constant. So the point is the following. So conversely, under the assumptions in green, there exists an holomorphic function determined up to a purely imaginary constant such that this is holomorphic. So since v is determined up to a real constant, f is determined up to an imaginary constant. What does it mean, this? What does it mean? It means that, well, this is the theory of one-forms. So do you know what is one differential form in the definition of one-form? Do you know what it is? So if the name of one-form is new to you, you have to ignore this 10 minutes discussion. Don't worry. Doesn't matter. Doesn't matter. In any case, the definition of exact one-form, I recall it. But so if the one-form, say, omega is one-form in R2, yes, in R2, or in Rn also, it is called exact in omega. It's called exact in omega. So if it admits a potential, it admits a potential, which means exactly if there exists a function phi such that omega is not any one-form, but is a differential. But since the theory is a differential form, it's non-trivial. If you have not studied, don't worry. You can skip this part. This was just a remark, maybe an exercise. OK. Maybe another exercise that we can do. Let us consider the function in one dimension phi of x equal 1 half of x. And let us compute, so this is in one dimension, take, take, so this is exercise, take so-called test function. So take a function psi in C infinity with compact support in R, so-called test function, and compute the following quantity, U psi second in the x. OK. Compute this. OK. This is, let us compute it. So this is sometimes also, well, let us compute it. So this is equal to what? This is minus. So U is this, which is therefore, so we can use integration by parts here. And this is U psi prime dx. And then there is a boundary term, but this boundary term is what? It's 0. So now what is this? This is minus the integral over R minus the negative integral, the negative part of reals. So this is, this function is minus 1 half and plus 1 half. This is U prime. OK. Therefore, this is minus. Then we have. So let me compute the, let me put the minus here for simplicity, OK, minus, plus, plus. U prime is minus 1 half here. Therefore, it's minus 1 half psi prime plus dx plus 1 half psi prime dx. OK. Ah, sorry, this is phi. Excuse me. Phi. Sorry, sorry. Phi. This is phi prime. Thank you. Phi prime. Yes, but it is differentiable almost everywhere. So this is an integral which is respect to the Lebesgue measure and you will. OK. Then what is this? Sorry. So then this is compact support. So we can integrate by parts once more. So this is 1 half. So there is this minus 1 half. So this is, let's say, the integral between psi of 0 minus psi of minus infinity, which is 0, plus 1 half. And then we have minus 1 half, again, psi of 0. OK, so we have concluded that the integral of phi psi second r is exactly equal to psi of 0. For any psi infinity c, OK? This is just, for the moment, remember this as an exercise. For any test function, phi psi second equal psi of 0. OK. So now let me continue with the properties of harmonic functions. So as you can see from what is written here, the theory in two dimensions of harmonic function is very, very close, is strictly related to the theory of volumetric functions. However, in higher dimensions, we have not anymore that strict relation. So we now pass to more general results in any dimension. And so we will not use now the theory of volumetric functions. Let me also remember maybe this also was done. Let me also recall the second green identity. Also, this has been done last time. So maybe it is not necessary to rewrite it. So definition. So let omega be an open set of rn. We say that u is subharmonic, and let u in c2. We say that if u is subharmonic in omega, if minus Laplacian of u is less than or equal than 0 in omega, and we say that u is superharmonic in omega, if minus Laplace of u is larger than or equal than 0 in omega. And of course, it is harmonic if it is both at the same time. If it is subharmonic and superharmonic in omega. These are definitions. So then we have the following theorem. Maybe this is called weak maximum minimum principle. So let omega be bounded. Let omega be bounded. Let u be subharmonic in omega. Then the max of u over the boundary of omega. Sorry, subharmonic let u in c2 intersection c0 of the closure. Because in the definition of subharmonic, we just need c2. But now we want to consider also the values of u at the boundary. So it is important now to assume that we can speak about the trace of u on the boundary. So we assume that u is also continuous up to the boundary. So the max of u is equal to the max on the closure of u. So this is the maximum principle for subharmonic functions. It is the analog of the maximum principle for subcaloric function. I would say, I mean, we proved a similar statement for the heat equation. Where the operator was not minus laplacian, but the operator was ut minus laplacian. And in that case, we found that the maximum value was assumed on part of the boundary. Remember? So here, in place of this, in the parabolic case, there was this part of the boundary. So somehow, this, on the other hand, says that this is time. On the other hand, if there is no time, this says that the maximum on the boundary is the maximum value of the function. So in some sense, if you look at the parabolic problem, as where t were any variable, not special, but any variable, the parabolic weak maximum principle says slightly more than this. Because it excludes this part of the boundary. Now, we proved this result. We have already proven this. Remember, in the parabolic case, we have proven by constructing maybe a strict sub-solution. Remember? So in the parabolic case, we were reduced essentially to this. And now, we will do the same. The idea was to slightly perturb the function u. In the parabolic case, the idea was to define v, tx. Maybe it was u, tx, minus some content like this. Thank you. Thank you. Yes, the statement is still not complete. Now, these were comments. OK, so just to remember, the proof was based on reducing to the case of strict sub-solution and then proving the assertion. Passing to the limit as epsilon goes to 0. Yes, this is the weak maximum for sub-solution. Of course, there is the complete equivalent. Since now, the operator is you can change u into minus u, changing the inequality, changing minimum with maximum with minimum. So if you assume superharmonic here, then you exchange your minimum. This is the assertion for. So this is concerning the minimum principle. And of course, if the function is harmonic, then you have at the same time the two equalities. So the weak maximum and minimum principle at the same time. Now, again, why the word weak here? Well, this does not exclude that there is an interior maximum. It says only that the maximum value, it is assumed here, but it doesn't say that it is only assumed here. So in principle, you could have also another interior maximum for the moment. This is why we call it weak. So I think that the proof is quite similar to the previous case. So we can consider the function v of x equals, say, u of x plus epsilon e to maybe x1. If x1 is the first coordinate, epsilon, say, is positive. Let us try this. So what happens to the Laplacian? So the Laplacian of v of x is equal to the Laplacian of v of x plus epsilon e to the x1 again, right? OK. So minus Laplacian is minus, minus. OK. But we are assuming now the case of sub-solution. So this is by assumption less than, so this is for any, so this is a definition in omega bar. And when x is in omega, we know that this is non-positive by assumptions. Because we are assuming sub-solutions. OK. So for a sub-solution, this is less than or equal than 0 by assumption. OK. But this is strictly negative because epsilon is positive. Therefore, this is a strict sub-solution. This is a strict sub-solution in omega. So a slight perturbation of the sub-solution u gives a strict sub-solution. OK. Now, we claim that, now what happens? Assume that x0 is a maximum of v. Well, if it is a maximum of v of v and it is interior because omega is open, but omega, of course, is open. Sorry, I have not written, but it is always open. And bounded. Bounded is very important, because otherwise we don't have max here. And also, actually, the result is false. Remember, what happened for the just parenthesis? What happened for the heat equation? For the heat equation, we had the weak maximum and minimum principle on bounded domains. This gives uniqueness of the Cauchy problem, say, on bounded domains. But we know that on unbounded domains, this uniqueness, in general, is false. So something is happening on unbounded domains. This was the message. And the same is here. So for a bounded domain, we have this. And we will see that uniqueness actually, this is false also in unbounded domains. For instance, as a consequence, also, say, uniqueness is false in unbounded domains. So just a comment. So the theory in this case parallels very much the parabolic and elliptic case are, in some sense, very similar, at least under this point of view. Very, very different from the hyperbolic theory, the wave equations, which was completely different behavior. OK. So this is a strict subsolution. And therefore, assume for the moment that we had, if we have an interior maximum of v, then the Laplacian of v has a sign at the maximum. The Laplacian of v at the maximum is less than or equal. Right? Which is a contradiction with this. Because from here, we have that inside the domain, side omega, the Laplacian is always positive. So we have a contradiction. This means that concerning the function v, the maximum points of v are assumed on the boundary. So we have proven, for v, for a strict subsolution that x0, the maximum points of v are. So now take an arbitrary point x bar. So take now any arbitrary point x bar in omega. So this is now. So u of x bar, it is what? Well, u of x bar is, by definition, v of x bar minus this. v of x bar minus epsilon e to this x1 bar. This is u x bar. Thank you, sorry. u x bar. This is by the definition of u of v. And this is clearly less than, instead of v, I take the maximum of v itself, 1 omega bar. But this is equal. This is equal. Because we have here seen that the boundary for v. And this is now replaced. Now we want a search on concerning only u. So we replace u here in place of v. And so again, this is less than or equal than what? The max of u, because u v is this, plus the maximum of this on the boundary of omega, which is bounded by a constant times epsilon, clearly. Because this is x1 ranges in a bounded interval. And therefore, I take the maximum value of this interval of the exponential. So I have a constant here and an epsilon in front. At the end, this is, and I consider, maybe I can also take plus epsilon. So there is a constant times epsilon. There exists a positive constant. So now, letting epsilon to 0 plus, in this argument, gives that u at x bar is less than or equal than this max. And therefore, since now x bar is arbitrarily inside the domain, this immediately implies that the max of u on omega bar is less than or equal than the max of u. Do you agree? Since the opposite inequality is immediate, since the opposite inequality, this one, is immediate. Because this object is contained in this. The boundary is contained in the closure. We have the equality, OK? So we have a proof which parallels, say, the proof of the weak minimum principle in the parabolic case. Which are the corollaries of this? As I say, there is an interesting corollary. So corollary, weak comparison. So assume that u and v. So assume that omega is open and bounded, OK? u and v are in the right class. u is a sub-solution. v is a sub-solution, super-solution, perhaps. And u is less than or equal than v on the boundary. Then we can conclude the interesting assertion that this relation, or the relations, must be true also inside, OK? The proof, well, the proof is, so let us consider u minus v. Then u, OK? So maybe you can continue by yourself at this point. Because this is, by assumption, is sub-harmonic. So v minus Laplace v is less than or equal than 0 in omega. Because it is minus Laplace of u, which is non-positive. And that's minus, minus this. So plus this, which is, again, non-positive, OK? So this is in omega. And in addition, by assumption on the boundary of omega. Therefore, v, sorry, w is less than or equal than 0 in omega. Why? Well, because the maximum point of w, the maximum value of w, is the same as the maximum value of w on the boundary. And the maximum value of w on the boundary is less than or equal than 0. Therefore, also the maximum value everywhere is less than or equal than 0, which is equivalent to, say, this by the maximum principle, OK? So we have this interesting conclusion. And we have also the following remark. Maybe another conclusion, theorem. Again, I am sorry. Throughout this course, for the moment, we have never considered the existence problem. We have just either constructed a special solution of our PD or proved uniqueness once you have existence. So existence requires, usually, the notion of weak solutions, sobole space, and so much for the moment theorem. So let f be a continuous function on omega, say. And let phi be a continuous function. Boundary of omega, assume that omega is bounded, open bounded. Then the problem has at most one solution. So this is maybe the Poisson equation, called Poisson equation. And so we have a boundary value problem, minus Laplace in a view equal to f, u equal phi. And this is at most one solution, because proof, if u1 and u2 are two solutions, assumed by contradiction, that you assume that you have two solutions, then you want to prove that they are equal. So take the difference. The difference satisfies minus Laplace of u, minus Laplace of omega, equal to 0. The difference, of course, is in the right class of objects, and satisfies the Laplace equation in omega, and also it is 0 on the boundary. And so now we can apply the corollary. Because the corollary say, w satisfies both the maximum and the minimum principle. On the boundary is 0. Therefore, from the maximum principle, w is less than or equal than 0. And from the minimum principle, w is larger or equal than 0. At the same time, therefore, w is 0, but w is less than or equal than 0 in omega on the maximum principle, and w is larger or equal than 0 in omega from the minimum principle, unnecessary. So w is 0, which means that u1 is equal to u2. So we have uniqueness of the Poisson equation with the boundary conditions on bounded domains. Again, the remark is that there is no uniqueness in unbounded domains. So now we do not require a so difficult example like the Tykonoff one. You remember, the Tykonoff one is very clever. The Tykonoff example was very clever. The Tykonoff example was the construction of a function that satisfies the heat equation in the half space, which is 0 at the initial time, remember? But here is more easy. You can consider, for instance, the function on, say, take omega equal the half plane, positive half plane, say. And then define and take u of x, y equal to x. So this is an example. Then minus Laplace of u is equal to 0 in omega, obviously, because this is linear. So when you do two derivatives, this is 0. And u is equal to 0 on the boundary of omega, because the boundary of omega is y equal to 0. So this is a non-zero function, harmonic non-zero function in the half plane, 0 on the boundary. Of course, another solution of this problem is the function identically 0. So we have two solutions, so there is no uniqueness, because u is omega is unbounded. This is much more easy than in the Tykonoff case. Now, another remark, maybe. Can you imagine what does it mean, the characteristic vector for this operator? What could be the definition of characteristic vectors? The set of all characteristic vectors. Can you imagine what it is here? So remember, you have to take the principal part of the operator. This is all principal part. And do you remember in the case, say, of the heat operator which was the definition? It is the same. Do you remember the definition in the case of heat operator? In the case of heat operator, here there was n1 plus n, maybe. So c0, c1, cn. And then it was only the principal part, so c1 squared plus c2 squared plus cn squared equal to 0. Do you remember? And we deduced that in this case of the heat equation, planes parallel to the horizontal plane were all characteristic. This was the situation. Now, here, which is the definition now? c is equal, however. c1, c2, cn. There is no c0. So now c1 squared plus plus cn squared equal to 0. Yes, it is empty. Therefore, there are no characteristic vectors in the case of the Laplace equation. This means that if you take, for instance, a plane, then this is non-characteristic. So there are no characteristic vectors. So for instance, this is non-characteristic. So this means that if you take, this is say, x and y. And now take this as a, so if you look at the Cauchy problem, now you do a strange thing. I mean, you think as y as time, very strange. Then you look at the Laplace equation, writing the second of u with respect to the y squared equal minus du squared. And you interpret this as time. So and you impose to this problem an initial condition, u0 equal u bar. And du over dy in 0 equal to u1 bar. These are analytic. If I assume that this is analytic, this is analytic. This is a plane, so it is analytic. Then by the Cauchy-Kowalewski theorem, you surely have a local solution. However, this is not very natural because in general these are problems which are static. I mean, the Laplace equation has no time in some sense. There is no time. So it is not so natural to look at this variable as a time. I mean, it is not natural to solve a Cauchy problem for the Laplace equation. So even if this is a solution at the end in the applications for elliptic equations is not very useful. Not very useful. Analytic, analytic, not useful in elliptic equations. So now we pass to, well, before starting the part of functional analysis, I think that we have to say something more on harmonic functions, which are so important because they are so important in general and also in physics that we have to say something more. And so let me try to prove the following result. So let u in C2 be subharmonic in omega. Be subharmonic in omega. Take any point. Take a point in the interior. Now omega is just open. Omega is open now. Omega, open. C2, subharmonic, an interior point. And take rho positive so that, small enough, the open ball centered at x0 with radius rho is, say, far from the boundary. Oh, my god. So take a positive radius. Since x0 is interior, surely there is a ball with this point. Now consider the following. Now then, d over d rho of this, of u, is bigger. So if you take the notation, b rho of x0, this means the mean value. So it is the volume of the ball of radius rho, which goes as rho to the n up to a constant. This is the volume of the ball of radius 1, remember. So by scaling, the volume of the ball of radius rho is this, times u of y, d1. This is a definition. So this means that, then you have the similar inequality for a superharmonic function with less than or equal than 0. In particular, for an harmonic function, if this is true, in particular, if u is harmonic, then the map function, which takes rho into this, is constant. This is a very strong assertion. It says that the value, now, which constant is it? Is the value of u at which point? At x0, at the center. Simply pass to the limit in this. This constant equal to u of x0. So for an harmonic function, there is this very strong assertion that if you take a point x0 and you take a small ball around x0, the value of the function at the center is equal to the mean. You take the mean of the function on the whole ball. They are equal. This is a solid ball. It is a solid ball. Of course, in two dimensions, this reminds a little bit the Cauchy formula. The Cauchy formula says if you have an holomorphic function, then you can write the value of f at the point as a mean value of something around a circle on the boundary of a ball, on the boundary of a circle. Well, there is some difference, because this is, for the moment, is a solid ball. And this result is valid in any dimension. Of course, in two dimensions, it's strictly related to that one. But in any dimension, this is true. This is a consequence. And this assertion here will have, in turn, a lot of consequences. So maybe it is worthwhile to prove another consequence of this remark. Consequence, maybe before proving this formula, which is called maybe the mean value property. Yes, the mean value property. This is called the mean value property. The mean value property says that this is true. And then let us see, before proving this, let us see a consequence, an immediate consequence. So theorem, strong maximum and minimum principle for harmonic function. Strong now for the first time. It is strong. So it says the following. Assume, for simplicity, that omega is connected. Otherwise, you reason connected components for separately on any connected component. So assume that omega is connected. If x0 in the interior is a maximum point of u, then u is constant. And the same for the minimum. Same for the minimum. So if you have an interior minimum, then it is constant. So this result now excludes what was left open by the weak maximum and minimum principle. But the weak maximum and minimum principle was saying, OK, I don't know. Surely, I mean, for instance, for a sub-solution, the maximum are assumed at the boundary, surely. So if this is the maximum, maybe it may happen something like this. But the maximum are assumed at the boundary. But this does not exclude that there is another maximum point in the interior. This was the content of the maximum principle. This says that if this happens, then actually, the function is constant. So necessarily, if it is non-constant, you have this picture. So it's a strong qualitative property on the graph of the solution knowing only something on the Laplacian. Because, of course, for a convex function or a concave function, some of these results are much more understandable. The point is that here, you're not assuming an equation on the action. You're assuming just an equation on the trace just only on the Laplacian, but nevertheless, assuming an information only on the Laplacian gives such a kind of result. So let us see the proof of this. Let us see the proof. Well, assume that you have x0. So it is enough to prove that define m as the max of u take omega also bounded. Sumo can connect it and bound it for simplicity. And define m as the maximum of u on omega bar. Then the set of all points x in omega such that u of x is equal to m, the claim is that this set of points is at the same time open and closed in omega. If I prove that this set is at the same time closed, my closed is obvious. But open also in omega, since omega is connected, then necessarily this must be equal to omega. And therefore, u is constant equal to its maximum value. So it is immediate to see that this is closed because u is continuous, so closed. Yeah, I'm assuming that omega say is bounded, so it admit the maximum. If all maximum points are on the boundary, there is nothing to prove. It is enough now to show that it is open. And to show that it is open, we use the following. Something missing. What do you mean? If this is harmonic. Yeah, but I mean it's constant. And therefore, if you put here omega n, then the constant will slightly change. Yes, this is actually constant. It's independent of everything. You're right. If you want exactly the value at the center, you have to put omega n. Otherwise, if you don't put omega n, this is still constant equal to omega n. Thank you. So let us prove that this is open. OK, take a ball. So take a point x such that, take a point x not in omega, such in omega, such that u of x not is equal to m. Take a small ball centered at x not small enough such that it is strictly contained in omega. And then this is constant. Then 1 over rho to the n. So the mean value of u is constant, independent of rho. Rho, this function rho that goes to the mean value of u on the solid ball. This is constant. But u in this ball, but u is less than or equal than m in the ball because m is the maximum. Therefore, u is less than or equal than m in the ball. So how is it possible that u at the center is equal? So the maximum value at the center is equal to this mean value. And at the same time, that u is less than or equal than the maximum in the solid ball. Necessarily, u must be constant, right? Otherwise, it is impossible. I repeat. I know that u at the center is its maximum value. And I know also that this maximum value is the mean value of u on a ball. But I know also that on a ball, u is less than or equal than m. So therefore, I conclude that necessarily u must be equal to m in the ball. Is it clear? It's immediate. So but this shows that this ball is contained in this set. So if you have a point in this set, there is a wall ball inside this set, which means that this set is open. And so this shows that it is open and closed. And so we have this very strong qualitative property. So I have to prove this. And this, in turn, this strong maximum and minimum principle has also many other consequences. By the way, concerning the heat equation, there is, again, there is also a strong maximum principle. And one can prove it using also in the heat equation case, there is a mean value property. But it is much more difficult than this. So maybe this is, again, another remark. Also in the case of the heat equation, there is a mean value property. The point is that but b rho of x naught is replaced by much by strange, I strange, strange ball. However, there is such a kind of formula. And from that kind of formula, you can, again, prove the strong maximum principle. OK? So as a consequence, you also, in the case of the heat equation, you have that, necessarily, the maximum and the minimum point is on that strange part of the boundary. There cannot be interior. If you want to look at this curious mean value property, for the heat equation, you can look at the book of Evans, for instance, in the Evans book. OK. So now let us try to prove this, define function phi of x, which is this form, define this function here, OK? And consider, so phi of x, maybe let me call rho, sorry, rho squared. And take x naught for simplicity equal the origin, OK? So this is actually the same rho. And take x naught equal the origin for simplicity. Now, what are the properties of this function? So phi is equal to 0 on the boundary of the ball. Because on the boundary of the ball, the norm of x, the boundary of the ball, centered at the origin of radius rho on the boundary, this is 0. So, OK? Also, it is positive in the interior, OK? So just a paraboloid, say, vanishing exactly on the boundary of the ball, OK? So therefore, I have that, now I'm integrating on the ball, phi Laplace of u. And this I know the following. Phi is positive in the ball. And minus Laplacian is less than or equal than 0, because it's a sub-solution, it's sub-harmonic, OK? Therefore, plus Laplacian is larger than or equal than 0, and phi is positive, OK? Therefore, this has a sign. Now I apply the second green identity. So let me rewrite the second green identity. So the second green identity, all the following, remember, was let me call, say, u and v here. v Laplace of u minus u Laplace of v is equal to v du over the nu, nu is exterior, minus u dv over the nu on the boundary, OK? This was the second green identity. So now I apply this formula with phi equal to v. So the integral over n omega equal to b rho. So b rho phi Laplace of u dy is equal now to u Laplace. So let us compute the Laplace of phi here. So it is equal to, so let me compute the Laplacian of phi. So the gradient of phi is equal to minus x i. If you differentiate with respect to x i this equality, this, of course, is a constant. 1 half cancels with these two. And then it is clear that the gradient of the norm square is just the identity map. It's just x. Therefore, the Laplace of phi is equal to what? It is equal to? Is equal to? No, it's not 0. It's minus? Well, we have 100 variables, minus n. Minus n, number of variables, OK? Because the Hessian of this is just minus the identity matrix. And if you take the trace of the minus the identity matrix, the identity is n times n, and so you sum 1 n times, OK? So this is equal to minus n. Do you agree? Is it OK? So let us consider now this. So this is equal to this plus this. Then I have n times u in dy. And then I also know by this remark that phi is a function which is 0 on the boundary of the ball. Therefore, this is 0, OK? Because u is equal to phi. I'm taking u equal to phi, v equal to phi. So I'm taking v equal to phi, OK? If v is equal to phi, v is 0 on the boundary of the ball. So this term is not present. So we have only this, 0 u d phi over d nu, OK? And we know that this is larger than or equal to 0, OK? Now I have to compute the phi over d nu. So I have already written that the gradient of phi, so this is equal to n integral on the ball u dy minus u. Now, the gradient of phi is minus the identity, minus x. Therefore, this seems to be plus scalar product dot nu, d h and minus 1. If I make some mistake, please correct me immediately. So now who is nu? The exterior normal is what? It's x over rho, right? Because it is exterior. So this is, so nu is x over rho. So this is the scalar product of x with x over rho. And so at the end, it's equal to rho. Do you agree, please? n integral over b rho of u in the y plus rho integral over b rho u d h and minus 1, OK? It seems that there is a mistake because there is minus n. So there is a minus n here. And there is a minus n here. Sorry, there is a mistake. Because I have u Laplace of phi on the right-hand side, and Laplace of phi is a minus n. So let us look at the inequality that we have obtained. So minus n, the integral over b rho of u plus rho, the integral with the boundary of b rho of u is larger than or equal to 0. Let me check that I am correct. Yes, yes, yes. So this is the inequality that we have. And therefore, we can write it as follows. u less than or equal to rho integral over b rho. This is our inequality. And now let us compute that derivative, OK? So now let us compute the derivative of that map. So the derivative of 1 over rho, the integral of b rho of u. What is this? So this is minus n over rho to the n plus 1, the integral over b rho u dy. So this is a product, the derivative of a product. So I differentiate this. And then now what is the derivative? Can you imagine what is this or not? Can you imagine what is this? It is not difficult. I mean, you have to integrate on this ball. You can b rho. You can take the differential, different quotients, for instance, b rho plus h, the integral over b rho plus h minus the integral over b rho. Therefore, you are integrated, actually, on this circular annulus in any dimension, of course, in the differential. I am discretizing this derivative, OK? This is almost the integral over b rho plus h minus the integral over b rho divided by h. And h is exactly this. This is h. Therefore, if I now integrate the functions only on this object here, this circular annulus, divided by the height, what is reasonable is that this is equal to, converges to what? The integral on the surface. So indeed, this is equal to minus n. So I rewrite the first addendum, b rho, u dy, plus 1 over rho to the n, the integral of the boundary of b rho of u in the h n minus 1. And this now, this look, let's take 1 over rho to the n in front of everything, minus n over rho, integral over b rho, plus the integral boundary of b rho. And you see immediately that this, now, OK, this, yes, it is exactly by this inequality is immediately larger than or equal to 0. Because if you simply divide by rho, rho is positive, you can write this inequality equivalently like this. And therefore, this minus this is larger than or equal than 0. So this minus this is larger than or equal to 0. And this concludes the proof. Of this mean value inequality, which has a lot of consequences in the theory of harmonic functions.