 Zdaj mi se odličiti, da smo prišli, da je omega in u, c2, omega in u je harmonijna v omega. the we have shown that for any x0 in omega for any rho positive so that the bowl of radius rho centret at x0 is contained in omega then you satisfies the mean value property, u of x not, u歪-de-歪. So this condition can be also written as follows, the distance of x not from the boundary of omega is larger than rho. V što je vsega površenja, vzivaj smo inočnje vsega. Zdaj je x0, zdaj je izgleda vzivaj vzivaj x0 in vzivaj vzivaj vzivaj omega. Zdaj, na toj površenji razkaj je, da je to. In na zelo, z vzivaj vzivaj x0, to je rho. Kaj sem komputenja, da je to zelo, Tukaj je to zelo, da je 1 divided omega n rho to n integral. To je vzela, da je vzela. Tukaj je to zelo. Tukaj je, da je vzela, da je integral vzela, da je vzela, da je vzela. In nekaj notčenje, da je poslut, da vsezajte, da je taj, taj je u, taj je vsezaj, taj, taj je zelo, zelo vizajte, taj volum vsezaj, rečenje radii, zelo vsezajte, taj je vsezajte, OK? Tako za vsezajte, da u in za zelo zelo vsezajte, to je bilo, da smo, da smo, je bi izvrs àn za solid-buljeinnji. Mi重ini že vsi squadu občinujeva, su da su tamšelji nad bin, in da je volaitiasne odlič knowle do Lake. Nič je goto flashing. Ja bi mo chính malo ... ali če da se zelo prišli na zelo. Tako, če to je začelovano do zelo. Zato vse malo otvrili, da srečno so nalib smo drugi povod, ne? Moj bo se na taj subarmonijkičnjih vse. To nekaj, da se nekaj dobrote, čekaj na pravdje, ali, da je to vzeločno, da je svojo neč del, neč nekaj dve, da je inakvalitet, kar je. Tako, na pravdje tudi je teorema. Čekaj, da se prišli, da je to način, da je tudi konditon equivalent na odlično. Mene se počučili, da je harmonija, če je tudi to vzela. Tukaj smo imeli zvom, da je to tudi vzela, da počutek je, kako jim dobro tega je, da je in vzela, če je to kaj je vzela, kako je tudi vzela, zelo se vzela, da je to na njim rešenem, na njim fisku, na njim x0. Če je to tudi harmonija? Ne zelo. Ne zelo. Poznačaj. Daj sem zelo poznačil na taj dobroz. Srečno, da bo pomečila nešto boljške vseh. Nezelo. Poznačila, da jasne? Uvaj, da je vseh vseh lepšča. in pričelega, da je... naredajša je za nedaj, da je... zame, da ja izpravim, nekaj. Tih, tih, tih tih tizi. Zame, da... zame izpravim, nekaj s tem보je. Vse vične teže, da taj vsi izpravim, je to različno najgratne glasice. Zato bomo našlično vsečenje, ki se nalidem. Zelo je, da je to, da bomo vsečenje. Zelo je, da bomo vsečenje. Našlično je to, da bomo vsečenje. tj. tj. tj. u je v zelo, dan, u je vse vse delovate. Zelo taj gradječ je vse, kaj je vse zelo zelo zelo zelo zelo, In tudi, nekaj bilo problemi v zelo, kaj je zelo. Kaj je taj teorem? Zelo, da sem izgleda taj teorem. Zelo, da sem izgleda taj teorem 1 in taj teorem 2. Zelo, da sem izgleda teorem 1 in teorem 2. Zelo, da sem izgleda taj teorem 1, zelo jaكرere, da charge n gu Messiah, neko, da, ceh čutih čutih čustih, neko zelo tanem, da taj teorem test početsub glueva, da парim na začinку koptype. Zelo, da je stvarte angole parom do temnev vek pač preko ne rangedivn Clinic, bojo tudi in zelo makir. In ne da se naredite, nekaj je do svoj naša好好ja otoga? Ovo, tako, tko se bo v štom, in vs. c2 je, ki jih septimo, tako tudi je harnessa v klas, Čekaj, kaj je tudi tudi početil, da je tudi početil, in kaj je tudi početil, da je tudi početil, In ki pa vse blinki nekaj, bral sem, ok .. bral sem na vse druh sekolov, na vse druh sekolov, 3, 5, 4. Druga je, da sem na vse priredena bez zrbunje njal poest, da sem na vse priredena bez zrbunje njal poest, ok. to je, da, if this assumption holds, then this assumption holds. Therefore, it is clear that theorem 2, if I prove theorem 2, I prove sort of stronger version than theorem 1. Simply because the assumption of theorem 1 implies the assumption of theorem 2. It is stronger. So, actually, therefore, theorem 2 is stronger than theorem 1. Let us try to prove it. Remembering that, we can never write explicitly Laplacian of U, because this is not possible. OK, so, let us take, so, these are called test function. You see, this assumption is an infinite number of hypothesis, because for any test, this must be true. And, of course, the test has an infinite number. So, this assumption actually is an infinite number of assumptions. Now, we have to produce, we want to show this assertion, that we have to produce a clever choice of test phi. So, fix, so, proof of theorem 2, fix x naught omega, rho positive, rho less than the distance, rho less the distance from the boundary. And now, choose, define phi of x 1 half rho square minus x minus n naught square. This is the definition, OK? And then, actually, sorry, this is not enough, maximum with zero. So, it is, this means, what does it mean, this symbol? It is the max between this number and zero. Simply a symbol. What is this function here? OK, this is a sort of object. So, if I have x naught here, then this is positive till the distance. So, then I take the bowl of radius rho centered at x naught. This is rho, sorry, this is rho. When this is less than rho, then this is positive, and then it becomes negative. But then I take the maximum with zero, sort of function like this, OK? So, this function phi is concentrated on the bowl, and then in zero else. In particular, it has compact support in omega, because the bowl is strictly inside omega, and therefore the test function has compact support in omega by this assumption here. Is this clear or not? So, phi has compact support, and it is also lip sheets. It is not C1, of course, but it is lip sheets. Namely, what it is true is that the slope here of phi is finite. See? It is sort of object like this, and here the slope is finite. It does not happen in a situation like this, where the slope is infinite. This is not lip sheets. The slope of the gradient is infinite, but here the gradient, this is a sort of object like this, and then tronkated. So, where I tronkate, I am arriving like this, and not like this. It is clear by the definition of... Therefore, this function phi is admissible. I can put phi here. So, phi is a test, so-called test function admissible test function. Namely, phi is in the class. So, why I am taking this phi? Well, first of all, I want to have something concentrated on the bowl. Therefore, it is natural to take something which is zero out of the bowl. Next, essentially, this is a radial test function. In some sense, it depends. It is sort of radial symmetry. So, maybe it is the most natural choice of phi that one can do. So, now we can... So, remember, we are proving theorem 2. Therefore, this is our hypothesis. Phi, we have chosen one possible phi. Now, see what happens when I put phi here. That's special phi. So, let me raise... So, inserting this phi in this equality, we have... Now, phi... Another property of this leap sheets function is that grad phi. Since phi is leap sheets, it is almost everywhere. Moreover, it is zero out of the bowl. Simply because phi is constant. So, actually, this integral here concentrate on the bowl. So, this is equal to this and this is zero. Therefore, we have this. Now, let us see... So, let us compute the gradient of phi separately. So, the gradient of phi at x is equal to is the identity, essentially, x minus x naught. And therefore, we have the following grad u at x, x minus x naught. And I claim that from this equality and I claim that from this equality I should be able to deduce the assertion. So, let us see what can I do here. Remember, my u is only leap sheets. So, I cannot differentiate it twice. But this is a gradient. So, I can integrate by parts, putting the gradient on this side. Okay? Integrating by parts. What's happening? First of all, I have a minus u of x. And then I have the divergence of this vector field. Which is the divergence of this vector field plus the integral. Then I have... Sorry, let me rewrite for you the divergence formula. So, u eta on, say, an open set is minus divergence of u eta on an open set is the integral over a of grad u scalar product eta plus u divergence of eta. But this is also equal to u eta So, I am using the standard version of the... So, now, I have this. So, now, a is the ball, eta is x minus x naught. Therefore, this is equal to minus u divergence of minus this, which is written here, plus u and then I have the scalar product between x minus x naught and normal to the ball. Outward normal to the ball. Okay? Now, let me compute... So, now let me compute the divergence of that vector field. Do you know what is the divergence of this vector field here? It is equal to... So, it is equal to minus... The divergence of the vector field x minus x naught, the identity. Remember, the divergence of any vector field is the sum of the... of this. And constant... Is a constant divergence vector field and this is equal exactly to the dimension. Okay? Minus n integral over b rho of x naught, u of x dx. And then I have also that quantity there. Now, the... The outward unit normal to the ball. Do you know? What is this? Divided by... Okay? This is x minus x naught divided by rho. Okay? So, this is the unit normal and therefore this becomes... I have the scalar product so this actually becomes... So, the scalar product between this and this. It is just rho. U of y. Is it okay? Here, let us compute now d over d rho separately. Let us compute this. So, what happens here? You see, there is a volume integral and the surface of u and the surface integral of u but multiplied by rho. So, now, if I compute this I have minus n omega n rho to the n plus 1 v rho over x naught u of y dx plus rho to the n and you already know that... So, sometimes I write the variable integration x or y is of course the same. Sorry, I can write it x if you want. It is of course the same. Okay? Therefore, this is zero and so minus n now if I divide it by rho to the n plus 1 1 over omega n I find so that this equal to zero implies that this is equal to zero. Do you see it? So, from this equality equal to zero it follows that this derivative is equal to zero. Hence, this means that this quantity is constant is independent of rho. Okay? Since u is lep, she said that therefore is continuous I can let rho go to zero and therefore u at the center is... So, this constant... So, let me write it, maybe it is better. So, one... So, this is constant independent of rho. Now u is continuous therefore the value at the center is equal to the limit of this. Okay, so we have proven a slightly stronger version theorem 2 you see is slightly stronger than theorem 1. Now, assume that you want to... So, I leave you this as an exercise now. Assume that you want to simulate a similar statement, but for sub-solutions. Okay, so for sub-harmonic functions. So, assume that you want to modify this statement for sub-solution. You would like to have as assumption this but you don't have it. You don't have it because u is just lep, she said not c2. So, you want to write this in a weak form like this. How can you write this in a weak form? So, we have already observed that this is a weak way to write minus laplace of u equal to 0. So, it is a natural way without writing to derivatives of u. Do you agree? Now, the point is which is the weak way to write this condition. Say, u lep, she is now in this condition. Okay. How can I write this in a weak form? Exercise. So, you see, what happens? Let us try to reason as before. Let us multiply by a test function and then integrate by parts. So, now, is this true? No, when it is true? Fine. Therefore, we have already the answer to our... because if phi is positive, this is still true. Then, when I integrate by parts, I have to put the minus in front of this minus, which becomes a plus. And therefore, grad u grad phi less than or equal to 0 for any non-negative phi. This is a weak... So, this less than or equal to 0 for any non-negative phi in this class is the weak form, the weak formulation of this condition. Do you agree? Or not? So, this... Okay. No. Do you agree? That's okay. Therefore, this is the weak natural form of writing this inequality without never writing two derivatives of u. And so, the exercise homework is... This now is the assumption, which is the thesis. Can we deduce that u of x naught say is less than or equal than the mean value? Can we deduce something like this? Try to do by yourself a similar proof and see what happens. Okay. Now, another exercise maybe that we do together. So, assume that u now is just only continuous than u. Satisfaj the mean value property in the volume sense. So, for any x naught, for any rho, et cetera, et cetera. The previous exercise is take an open set omega over m. Take a Lipschitz function u and assume that this is true. So, for any Lipschitz test function non-negative, this is less than or equal than zero. Then, of course, you cannot deduce that this is equal to this. But maybe you can deduce that this is less than or equal than this, or larger than or equal. So, try to find the right inequality. Try to prove, for instance, this. What you have to do is exactly the same proof as before. Notice that the previous proof, our test was this, which is non-negative. Therefore, you can make exactly the same proof, because this test is allowed. So, you have exactly to follow the previous reasoning. And you can use the same test function, because this is outside this. So, sorry, here it is for any x0 in omega, for any rho with the distance less than etc. So, write correctly. This is equivalent for any boon. It is equivalent to say that u of x0 is equal for any. So, the mean value property on solid bowls is equivalent for a continuous function. Is equivalent to the mean value property on the hypersurface, on the boundary of the bowls. So, this means, I recall, that this means 1 over omega n rho to dn by definition. This means 1 over I think this the boundary integral and this is a surface integral. So, instead that integrating here and divided by the volume you integrate just only here and divide by the area. So, actually if this exercise is true then we have two possible ways of writing the mean value formula. Depending on what you prefer. There is just one point here that you see, for writing this kind of mean value formula in principle you just need that u is l1 log, because you integrate if your function is locally integrable then you can integrate locally around the bowl. But if your function is a in l1 log this is not defined. Because an l1 log function is defined almost everywhere and therefore there is no sense of the trace of u on an hypersurface. A zero labor measure. The hypersurface is zero labor measure. Therefore there is no way to give a meaning of an l1 log function on an hypersurface. What it is not defined is the trace of u on this hypersurface. Because this hypersurface has zero labor measure and you can assign whatever value you want remaining inside the l1 log plus. So this is not well defined. This integral is not well defined. I mean if you have an l1 log function in the plane say or if you have an l1 log function on the line you cannot say and then you have a ball so this is a ball on the line. Of course you can integrate u here but you cannot say which is the value of u here on the boundary or the value of u here on the boundary. This is not defined. You cannot define the values of u on the boundary. You can always change whatever value you want remaining inside the back class. This is essentially the difference and so in order to not to have problems with this definition we assume at least continuity. Of course a continuous function leaves a trace on an hypersurface. So let us see the proof of this exercise. Assume for instance that your function satisfies the mean value property in the sense of surfaces and consider and so this is our assumption and I want to prove this. So now I start from this integral and therefore I write it so this is equal to this now it is clear I make the integral in polar coordinates. So this is equal to now I write this in polar coordinates so this is equal to the integral from zero to rho and then I have what do I have? I have the integral on the ball of radius r say x0 u of y dr this is double integral so this is dr also dr Now our assumption is this so this actually is a mean value on a surface of radius r therefore our assumption says that omega n-1 r to the n-1 u of x0 is equal to u of y d h-1 y so u of x0 is the mean value on this ball, on the surface of this ball by assumption so now I can substitute inside this integral this so this is actually equal to 1 over omega n rho to the n then I substitute inside the integral and this is a constant and this is also another constant so omega n-1 u of x0 and then the integral from 0 to rho to r n-1 dr hoping that I am doing the correct computation so what I find so this I can compute so this is equal to omega n-1 divided by omega n then I have u of x0 and then this integral is what is 1 over n rho to the n rho to the n so this seems to be a constant u of x0 but this is equal to 1 I was starting from the volume so I have shown that this is equal to this from this therefore we have shown that if u satisfies the mean value formula on surfaces of the balls then it satisfies the mean value formula inside the balls because at the end we have shown that this is equal to this so we have proven such implication now let us see the other implication so assume that for any x0 for any balls strictly contained in omega this is true and then I want to show that then this is true do you have suggestions for doing this so my starting point is this red box is equal to this so I know that this is true and I want to find something on the surface which could be the idea remember that we want to pass from a volume integral to a surface integral and there is a way to do this and the way is to differentiate so actually the left hand side is independent of rho therefore if I differentiate this with respect to rho must be 0 we have already done this computation so we have 1 omega n minus n sorry rho to the n plus 1 the integral over b rho u of y dy plus 1 over omega n rho to the n the integral over boundary of b rho u of y so 0 is equal to this now what do I do well therefore this is equal to this so let me write it as follows so minus so sorry I find that the integral over b rho u of y is equal to n over rho times this integral times this integral ok and this should be all because now I divided by omega n minus 1 rho to the n minus 1 rho to the n and then I have omega n minus 1 and again omega n 1 over omega n is n over so this is equal is it ok? so it is interesting to have at our disposal therefore 2 possible mean value properties for continuous function this is equivalent to this ok now let us do this let us do this following this theorem another interesting theorem here in the direction of the question so theorem 3 let you see omega satisfy the mean value property mean value property mvp now by mean value property is one of the two because for continuous functions they are equivalent so this is mvp then u is so what is the situation for the moment so uc2 minus laplace of u equal to 0 u and lip minus laplace of u equal to 0 in the weak form implies mean value property now we are showing that u continuous mean value property implies uc infinity in particular c2 and then we will prove this will imply that actually u is harmonic in the classical sense ok so let us see the proof of this theorem so let us choose let me denote it by eta epsion standard standard mollifiers what does it mean standard it means that they are radially symmetric radially symmetric it means that they are non negative it means that they are c infinity c infinity it means that they are the support of eta epsilon is contained in b epsilon 0 for instance the usual way is maybe eta of x equal e minus x2 if less than equal to 1 in 0 else maybe let me call this eta tilde and then eta of x 1 over the integral of eta tilde ok so it is simply a smooth infinity function and then there is eta epsilon of x epsilon to d minus x over epsilon ok so that we always have that we always have this normalization ok so standard sequence of mollifiers ok let us consider the function u epsilon for any x in omega such that the distance of x from the boundary of omega is larger than epsilon so I have omega here then I have all points so epsilon is very small actually so this is so say this is epsilon so I take a point here the distance of which from the boundary is larger than epsilon so let me call this maybe this set omega epsilon minus a little bit smaller than omega of the factor epsilon ok so for any namely for any x in omega minus epsilon I can consider the following the convolution of u with eta epsilon u convol ok so this is the integral over let me use the same symbols u of y eta epsilon x minus y by y y so this is the definition of convolution why I am taking x in this set well because it is obvious that if I take x here for instance then I am involving values of so I am actually integrating so the convolution is like integrating in a small ball centered x of radius epsilon so I go outside omega and u is not defined anymore ok therefore in order to have a meaning of this quantity I need to take x a little bit inside otherwise this integral is actually inside the ball of radius x ball centered x of radius epsilon and therefore I have problems for u so this is why but on the other hand for any x here this is well defined so u of x is equal to the integral over omega of u of y eta epsilon x minus y dy so now this is equal to what now let me recall who is eta epsilon so eta epsilon so at the end I want to show that this is equal to u so if I am able to show that u of epsilon is equal to u for any x in this smaller set then what I have proven is that u is infinity inside this smaller set because it is well known that u of epsilon because it is well known that this is infinity this is known ok so if I prove of course that u of epsilon is equal to u then I have that u is infinity in this set but then I can always take epsilon smaller and smaller so at the end at any interior point if I take epsilon small enough u would be equal to u epsilon in neighborhood of that point and u is infinity inside in the interior and therefore it is enough to show that this equality holds ok so let me compute this so this is equal to this is concentrated the support of this is just inside the bowl of radius epsilon centered at x therefore this is equal to u of y at epsilon of x minus y now it is radial ok so this is just an integral concentrated here at epsilon is radial ok now what is at epsilon remember that at epsilon of z is epsilon to the minus n eta of z over epsilon this is definition so we can construct just by commodity starting from just only one and we have this ok now what do I do so here now now I integrate in polar coordinates so that is equal to epsilon to the minus n integral from 0 to epsilon ok then integral over the boundary let me call this r maybe we call this r yes and then what do I have I have u of y dy and then I have eta of x minus y actually now x minus y is equal to r therefore I just have eta of r divided by epsilon and then I have u of y dh in minus y so I simply write this integral as the integral of 0 epsilon same quantity dh in minus y now I observe that when y is on the boundary of this ball this difference is actually r so since it is r is independent of y and therefore I can put it outside the inner integral ok now for the moment I have just used the definition of convolution I have used that the kernel is radially symmetric and supported in the ball that's what I have used now I use the mean value property because now my assumption is that u is continuous that is by the mean value property now I use it in the surface form I know that it is equivalent the volume and the surface form now I use it I see it from the proof it is for me convenient to use it in the surface form therefore I know what is this ok so this is equal 2 r to the n minus 1 omega n minus 1 u of x ok so I can substitute ah sorry thank you thanks so this is equal to this no with the dr with the dr sorry without the dr thank you now I can substitute and therefore what do I have so let me go maybe here so I use now the mean value property in the surface version u is continuous by assumption so I can do so I find epsilon to the minus n omega n minus 1 epsilon to the minus n u of x the integral from 0 to epsilon r to the n minus 1 eta over epsilon dr ok now homework by the property normalization property of eta we know that what do we know we know that the integral omega n on the ball of radius epsilon centered at 0 say of eta z dz eta epsilon z dz we know that this is equal to 1 remember we always have this eta equal to 1 and also this so I claim that this actually is equal to 1 check this at home if this is true then this is equal to u of x and this is true by our properties of normalization ok now I have erased my small scheme unfortunately so let us now show that the answer to our initial question so theorem let me call this theorem 3 maybe theorem 3 ok assume that u is c2 assume that u is continuous and u satisfies mean value property then u is harmonic and this statement is correct in the sense that in principle u is only continuous and therefore we cannot write Laplacian but we know that if u is continuous and satisfies the mean value property it is infinity so let me just add here actually there is also a remark that we made let me write it here that it is interesting remark we have already proven this remark actually if u is c2 and u is harmonic then u is infinity this is actually very interesting so this is a regularity first example of regularity theorem so regularity theory is maybe the most difficult part of partial differential equations it asserts that like this I mean you have functions satisfying a PD involving only two derivatives so it is rather surprising that if you look at this statement then actually u is infinity this is really surprising and this is I mean the point of regularity theory in PD is the most difficult part in theory is exactly this and why is this true because we know that in the proof we know how to do this u satisfies we know that if it is harmonic it satisfies the mean value property and if it satisfies the mean value property it is infinity so at the end it is interesting maybe theorem more than remark theorem it is a theorem first surprise so now so this says if we prove this this says that actually being harmonic or satisfying the mean value property is equivalent for continuous functions for proving this result let me introduce the function phi of r that is the mean value over the solid no maybe this again u of y dy let me compute this now let me define this for any r for any x for a point x in omega and r sufficiently small and therefore now we want to compute phi prime of r now to compute phi prime of r maybe it is better there is a difficulty here in computing phi prime of r because the dependence on r is on the surface so the idea is in order to overcome this problem here we are calculating on this make a change of variable and integrate it on the ball of radius 1 so that r goes inside here yeah but still let me let me compute this ok so I rewrite this as follows so I rewrite this as follows v prime do this u of x plus r z dhn minus 1 z so we know that this derivative has b0 ok but let me write it as follows so let me see if I am correct so if z is in the boundary of the ball then rz is in the boundary of the ball of radius r it is ok and then I am also translating the center now let me compute phi prime of r which is 0 at the end so this is equal now there is no r anymore neither here nor dividing by the unit volume unit surface area scalar product of grad u I already know that u is infinity so I can differentiate I am differentiating with respect to r therefore this is scalar product with z dhn minus 1 z ok and now I come back to the previous integral in the variable r so I come back to the boundary of br x grad u y and now z is z is this and therefore it is x minus y x minus y dhn minus 1 mistake y minus x integral of the ball of radius r centered x on the translated ball centered the origin of radius 1 so I do this first computation because for me it is easier to compute the derivative with respect to r now I come back to this and now what do I do this is you see what is this it is the scalar product of the gradient of u with the unit normal outward unit normal on the ball which we know to be equal now we have integration by parts ok so this is equal so let me divide it by the correct quantity r to dhn minus 1 and then I have now solid ball ok so this is equal to 0 because phi we know is dependent of r so phi prime is 0 ok ok so this is equal to 0 and therefore now I think that the proof should be almost concluded because since assumed by contradiction now so assume now that by contradiction that u is not harmonic so therefore there is a point where this Laplacian is non-zero therefore u is continuous actually it is c2 so assume for instance that this is positive at the point therefore by continuity there is a small ball where the Laplacian is positive which contradicts this therefore we have this interesting result that actually it is difficult you see but we have a sort of characterization of harmonicity even for continuous function if you have a continuous function which satisfies a mean value property then it is harmonic in the classical sense maybe I can leave you another homework very not so difficult perhaps consider the ball of radius rho center at the origin r2 so let call this omega ok take a function phi of theta equal cosine of n theta 0 to pi say and then find an harmonic extension so phi of theta is essentially defined on the boundary of the ball unit ball find an harmonic extension of phi in omega what does it mean that is equivalent to find anomorfic function function in the ball such that the restriction of its real part on the boundary so take for the ball of radius 1 for simplicity ok, such that find anomorfic function in the ball such that the restriction anomorfic function f such that the restriction of the real part of f on the boundary of omega is equal to phi this means that you are solving in the ball you equal phi and the same can be done sin and theta one of the two so it is not difficult not so difficult to find anomorfic function having this having this is not difficult can you imagine what it is theta is in no, actually yeah theta yes, theta is complex number of modulus one ok ok, theta is a complex number of modulus one but it is, I mean I mean, if you take f of z z to the n this is anomorfic and let us write it so if I write rho with now theta the argument so this is equal to theta now if I take the real part and I restrict it on the boundary of the sphere of the circle on the disk of radius one this is equal exactly to the cosine of n theta so in this sense theta is just the argument of the complex number so it is in between zero and two pi just the so maybe you can keep it like this so this is the answer to well, once you have this do this also for sinus then for the sum of two you can do because this is a linear function it gives you some example of harmonic function on the ball