 Welcome. In this lecture, we are going to discuss about mean value property and some of its consequences. The outline of this lecture is as follows. First, we introduce what is called mean value property and then we go on to show mean value principle namely mean value property holds for harmonic functions that is what is called mean value principle. And we show that functions having mean value property are C infinity functions and functions having mean value property are harmonic functions. In 3 and 4 points here, we are going to assume only continuity of the function. If a continuous function has mean value property automatically it follows it is C infinity. If a continuous function has mean value property then it is automatically a harmonic function. So, what is mean value property? So, we are going to define two notions of mean value property. They are called mean value property 1 and mean value property 2. First, let us define what is mean value property 1. Let omega inside R2 be an open domain. Let U from omega to R be a continuous function. The function U is said to possess mean value property 1 on omega if for every pair P, R such that the disk the closed disk of radius R with center at P is properly contained in omega. What does this mean? Take any point in omega and take any R positive such that this closed disk is contained in omega. Then the following equality holds U of xy, xy is P, P is P of xy. So, U of xy equal to integral of U over the disk DPR and this is pi R square is precisely the area of the disk. That means I am integrating the function on this domain DPR and dividing it with the area of DPR. Therefore, this is called mean value. So, U of xy is nothing but you take a disk around xy which is this DPR integrate and divide with 1 by pi R square. In other words, the value of U at the center of a disk is given by its average on the disk. Definition of mean value property 2. Let omega be once again an open domain in R2 and let U be a continuous function on omega. Then the function is said to possess mean value property 2 if for every point P in omega and for every R positive such that this closed disk is contained in omega something should happen. What should happen is U of xy is nothing but integrate U over the circle SPR denotes the circle center at P having radius R and divide with perimeter of the circle. In other words, this is the mean value of U on the circle is equal to the value of U at the center of the circle. So, this is called mean value property 2. So, we have a result which says that mean value properties 1 and 2 are equivalent. What does it mean? If you have a continuous function U on omega, then the following statements are equivalent. U has a mean value property 1 on omega is same as U has a mean value property 2 on omega. So, this means that if you have a continuous function and it has the mean value property 1, then automatically it has mean value property 2. Similarly, if the function U has a mean value property 2 on omega, then definitely it will also have the mean value property 1 on omega. So, what is the idea of the proof? We are going to prove that statement 2 implies statement 1. All the steps in this proof can be reversed that means statement 1 implies statement 2. So, let us prove 2 implies 1. So, let P be an arbitrary point in omega. Further, let R positive be such that the disk, the closed disk with center P radius R is inside omega. If U has the mean value property 2, then for every tau less than or equal to R, we have this equality. This is the mean on the circle equal to U at the center. Now, of course, by the definition it holds for R. Why does it hold for every tau less than or equal to R? It is because if D of P R is in omega, then D of P tau will also be a subset of omega for every tau less than or equal to R. Therefore, this holds for every tau less than or equal to R. Now, multiply the last equation with tau and then integrate with respect to tau over the interval 0 to R. So, I have simply multiplied this equation with tau and wrote this integral 0 to R on both sides. Now, we can integrate the left hand side because U of x y does not depend on tau. So, it is a constant. So, integral 0 to R of tau D tau that will give us tau square by 2 with limit 0 to R. So, that gives us R square by 2 into U of x y. Right hand side I have written as it is. Now, what is right hand side? It is nothing but the integral on the disc expressed in the polar coordinates. In polar coordinates tau D theta D tau is the area element and therefore, this integral is equal to this integral. Now, from here you get what is U of x y because you bring R square by 2 this side you get 2 by R square then you get 1 by pi R square into the integral U on the disc. So, that finishes the proof of mean value property 1. Now, how to retrace the steps and prove 1 implies 2. Let us do that. This is what we have. This is what we assume. If you have mean value property 1, we have this equation. That is on multiplying with R square by 2 we have R square by 2 U x y equal to this. Now, we can express this as this integral in the polar coordinates. Now, I want to pass from this equation to this equation. What do I do? Right hand side is the same. Left hand side is precisely this is nothing but this integral. So, we have got this. That means, we have this last equation on the slide. Now, here too I want to pass to this. How do I do that? Apply fundamental theorem of calculus. Then 0 to R will go away and what you will have is R times U of x y and here 1 by 2 pi 0 to 2 pi and tau replaced by R. Then we will have this. Then cancel R you have this. So, therefore, we can retrace the steps and prove that 1 implies 2. Remark in view of this lemma that we have just proved. We say that a continuous function has the mean value property on a domain omega, we do not mention which one. If U has either of the mean value properties, I am going to say that the function U has the mean value property. We need not mention which one of them because they are equivalent. Of course, in such a case U has both the mean value properties. If it has one mean value property, it has both the mean value properties. That is the reason we do not mention which one. So, the mean value property 1 may also be called the solid mean value property. This because in mean value property 1 features mean on the disc. That means I am integrating on the entire disc and taking the average. That is why it is called solid mean value property. And the mean value property 2 may be called as surface or circle mean value property. This terminology is not very standard. I am using this terminology because we can easily understand by names. Names represent a quantity much more than a very generic name like mean value property 1, 2, 3, etc. Solid mean value property means I understand it, how to write it. Similarly, surface mean value property, I know what does it mean. I can easily recall. Let us prove now mean value principle. So, let omega be a domain in R2 and U be a harmony function on omega. Then U has the mean value property on omega. So, proof of the theorem, we plan to show that U has a surface mean value property. So, if I want to show that U has a mean value property, I can choose to prove any one of them which is convenient. I am going to show that U has surface mean value property. So, let P naught be a point x naught y naught in omega be arbitrary. And let R be such that this open disk is contained in the closed disk that is contained in omega. And we need to prove that U at the center of the disk is given by the average of U on the circle. If U has to have mean value property, then the above equation namely this should hold when I replace this capital R with any small R which is less than R. This observation shows as a way how to proceed forward. So, thus we expect that for R between 0 and R, this equality will hold. In the above equation, the LHS is a real number U at the point x naught y naught is a number. Whereas the right hand side is a function of R, it depends on R. Therefore, the strategy is show that the right hand side is a constant function of the variable R. So, therefore, define a function V on this interval 0, R by this formula which is precisely the right hand side on the last equation in the previous slide. Suppose we show that dV by dr equal to 0, what does it mean? V is a constant function on this interval 0, R. So, as a consequence for every R in interval 0, R we get that V of R is constant, but what is that? That constant can be evaluated by taking limit rho goes to 0 of V of rho and that will be U of x naught y naught. On the other side, if it is a constant, V is a continuous function of R. Therefore, V of R is also equal to limit rho goes to capital R of V of rho and that will give you V of capital R which is precisely the right hand side in the circle mean value property. So, this completes the proof as V of R will be equal to U of x naught y naught. So, it reminds to show that this derivative of V with respect to R is 0. Let us compute. So, dV by dr is given by this. I have taken the differentiation inside the integral that is allowed because U is a smooth function, U is a C2 function. This is precisely this, this is the normal derivative of U on the boundary, that is normal derivative of U on the circle. Now, this is nothing but integral over the disc of Laplacian U dx and that is 0 because U is a harmonic function. So, we have used green's identity 1 in passing from this to this and harmonicity of U in concluding that this integral is 0 because the integrand is 0. So, if you have a continuous function which has mean value property, it is infinitely differentiable. So, let U be a continuous function on omega, omega is a subset of R2. Let U have the mean value property on omega, then U has continuous derivatives of all orders on omega and all the derivatives of U have the mean value property on omega. So, 0.1 says U is a C infinity function and this says all the derivatives have the mean value property on omega. Step 1, it is sufficient to show that the first order partial derivatives of U exist and are continuous. Once it is known that first order partial derivatives of U exist and are continuous, we can compute partial derivatives from this equation. This is known to hold because U has the mean value property. So, this holds. The question is whether I can differentiate this. I can, if I know that integrand is differentiable that is what we are establishing that first order partial derivatives of U exist and are continuous. Therefore, differentiating this equation is allowed. Note that the above equation holds for all R such that this closed disc with center at P is inside omega. As U satisfies mean value property on omega, where P is an arbitrary point in omega. Mean value property holds on omega means you take any disc, any closed disc which is contained in the domain omega, then mean value property that is U at the center of the disc is given by the average. Let it be a solid average or the surface average. So, the partial derivatives satisfy Ux of xy equal to this. I have differentiated with respect to x and similarly Uy. The above equations mean that Ux, Uy have the mean value property because this is exactly means that Ux have the mean value property. Repeating the arguments by replacing U with its partial derivatives, we conclude that U has partial derivatives of higher orders and all of them satisfy the mean value property. Now, let us show that first order derivatives in fact exist and are continuous. Since U has the mean value property, it has the mean value property 1 also. Therefore, we have this expression for U at the center of this disc. Now, writing the double integral, this is the integral on the disc, write it as iterated integrals, we get U of xy equal to, so this integral is precisely this integral. So, this equation tells us that Ux of xy exist. How? X is appearing here, right, inside and this integrand, imagine this is the first integral, then it has an integrand which is this, there the dependence on x is only through the limits and it is a nice dependence. So, by fundamental theorem of calculus, this is differentiable and we can compute. So, when you compute, when you differentiate this with respect to x, you get this particular integrand here. So, since the right hand side of this equation that is this is continuous function of xy, it follows that Ux is a continuous function of xy. So, thus we have shown that U has a partial derivative with respect to x and Ux is continuous function. Similarly, one can deduce the existence of Uy at each point and the continuity of Uy on omega. A continuous function having mean value property is indeed harmonic. So, let U be a continuous function on omega, assume that U also has a mean value property on omega, then U is harmonic in omega. It means Laplace in U equal to 0 holds on omega. So, proof of the theorem, we have proved this earlier, if U is a continuous function and has a mean value property, then all partial derivatives of U exist and all of them are continuous on omega, we have shown this. So, we want to show that Laplace in U equal to 0. Therefore, why we stated this is because Laplace in U should be meaningful because we have only started with a continuous function. So, as shown before, U has all derivatives in particular second order derivatives, therefore Laplace in U is meaningful and we can think of showing that to be equal to 0. So, on the contrary assume that Laplace in U is not equal to 0 in omega that means there is a point p0 in omega where Laplace in U is not 0. Now, Laplace in U is a combination of second order partial derivatives of U which are known to be continuous on omega. So, you have a continuous function namely Laplace in U which is not 0 at a point p0. Therefore, in a discard one p0, it will continue to be non-zero. Without loss of generality, let us assume Laplace in U of p0 is positive. What are the choices for Laplace in U of p0 when it is non-zero, it has to be positive or negative. So, let us assume it is positive. If it is negative, exactly the same proof, we can rewrite. Let us go ahead with Laplace in U of p0 as positive. By continuity of the second order derivatives, there is a disk D of positive radius epsilon centered at p0 such that Laplace in U is greater than 0 on the disk. So, we will be using the following identity in the computation that follows d by dr of U of this quantity by chain rule it is a gradient of U and then dot product with derivative of the inside quantity with respect to r namely cos theta sin theta. Also note that the outward unit normal to the circle at any point on it is given by x minus x0 comma y minus y0 by r. If you recall, we have said this earlier as dou U by dou nu at a point on the boundary of the domain that is at a point on the circle. This is the dou U by dou nu because this is a normal direction on the unit circle. So, for r less than epsilon, we have this that integral over the disk of Laplace in U is positive because Laplace in U at p0 was positive and hence we found a disk of radius epsilon on which Laplace in U is case positive and this by Green s identity is equal to this quantity and this is precisely this and r comes out because integral is with respect to theta and dou by dr also comes out what you have inside is precisely something related to mean value and therefore what you get is this quantity this integral is 2 pi U of x0 y0 which is 0 because this does not depend on r so derivative with respect to r is 0. So, on the last slide what we have got is 0 is less than 0 it is a contradiction. Our assumption that Laplace in U at some point is non-zero is wrong therefore U is a harmonic function. So, summarize today's lecture we have introduced two notions of mean value property and their equivalence was established we proved that harmonic functions have the mean value property. We have proved that continuous functions having mean value property are c infinity and continuous functions having mean value property are in fact harmonic functions. Thank you.