Dear students, hello. This is the sixth exercise class. Before coming to the actual solution
of the exercises, let me give you the statement of Young's Inequality for Convolution that
may mean probably you didn't see, but it might be useful to you. So it states that
if we have p and q in one infinity, any exponents, so not conjugate, f is in lp and g is in lq,
over n, then the convolution product between f and g is in lr, over n, for any r between
one and infinity, such that this equality is satisfied. So, sorry, let me rephrase for that
r between one and infinity, such that 1 over p plus 1 over q is equal to 1 plus 1 over r.
And the lr norm of f convolution of g is controlled by the product between the lp
norm of f and the lq norm of g. Coming now to the exercise sheet, as usually notation,
notation, notation, things we already know, so there was something new about it. So,
we recall the definition of Euler's beta and gamma function, where beta of xy is the
integral from 0 to 1 of t to the power x minus 1, 1 minus t to the power y minus 1 in dt for
x and y positive, and gamma of x is the integral from 0 to plus infinity of t to the power x
minus 1 e to the minus t in dt for x positive. And then, of course, these two functions can
be extended also for negative values of x and y, for complex values. There is only a nice theory
about it, but we don't need it. What you needed to know is that gamma, the gamma function is the
generalization of the factorial to non-integer values of x, in the sense that gamma of x plus 1
is equal to x gamma of x, so that n gamma of 1 is equal to 1, so that you can deduce easily that gamma
of n plus 1 is equal to n factorial, and gamma of 1 half is equal to square root of pi. And this is,
I mean, if you know the Gaussian integral, it's not difficult to check this fact. Instead, the
relation between the beta and the gamma function is given by this formula, beta of xy is equal to
gamma of x, gamma of y divided by gamma of x plus y. And this basically holds by making a, seeing the
gamma of x, gamma of y as a double integral and making some clever change of variables. But again,
you don't need it too short, just to tell you this thing can be done. Then we recall the fundamental
solution of the Laplace equations in Rn for n larger equal than 2, phi of x is equal to minus 1
over 2 pi logarithm of the norm of x if n is equal to 2, and 1 over n times n minus 2 omega n norm of
x to the 2 minus n if n is larger equal than 3. Remember, the fundamental solution is that
function differs, move in Rn minus the origin such that the depletion of phi is equal to the direct
delta centered in the origin and thus it is used to construct the green function over any type of
nice domain. And this was related to the second exercise. I noticed, moreover, that you have some
difficulties in applying Lebesgue dominated convergence theorem in order to ensure the
continuity or the differentiability of integral functions. So I thought it was, I'm sorry to
end with before, but I mean, this is a part of your prerequisites and knowledge that I think
you've already seen in the results you've already seen in courses of analysis, for instance. But
nevertheless, because they are straightforward, almost straightforward consequences of
Lebesgue dominated convergence theorem, but it's enough to know how do they work, which means
which kind of assumptions you need. So let us state here, take m and n integer natural numbers,
u and v open sets in Rn and Rm, f be a function from u times v into r such that the map which
sends x in fxy is measurable for all y and v and the map which sends y in fxy is continuous
for almost every x. So this means whenever you fix any y, the map is measurable. Whenever you
fix almost every x is continuous. Then you assume that there exists g in L1 of u such that
the absolute value of fxy is controlled by g of x for Lebesgue almost every x in u and all y and v.
Then f dot y is in L1, so as a function of x is in L1 of u for all y and v and the function
given by the integration of u in the x capital F is continuous over v. And the proof is
is really select y0 in v, then define the sequence fn which is given basically by fxny for any sequence
xn, sorry, fxyn, sorry, sorry, for any sequence yn converging to y0. This continuity property here,
which by the way holds for almost every x, that's enough because remember in Lebesgue
dominated convergence theorem you have that fn should, which I mean that your sequence should
converge almost everywhere, that's enough because it's an integral statement. So your new sequence
would be f of xyn. This is converging to fxy0 for almost every x because of the continuity here
assumption and it admits a summable measurement g because it's given here and so Lebesgue gives me
the continuity that means capital F of yn converges to capital F of y0. The other consequence is the
differentiation under the integration sign. So again the same assumptions beginning. Then we
ask that ff from u times v into r is such that the map which sends x in fxy is measurable for all y and v
and we assume that there exists y0 and v such that f dot y0 is l1 and
the gradient with respect to y of fxy the gradient on any any partial derivative of course
I wrote it for the gradient just for simplicity I was saying this this exists for almost every x
and all y the gradient y of course then you assume that there exists some function g in l1 such that
this gradient is controlled by g of x for almost every x and all y then we can deduce that f dot
y is l1 for all y of course one could here could ask already that it is always summable
but it is a consequence of this uniform bound so it's not necessary and the function given by the
integration over u of fxy is c1 over v and its gradient is equal to integral over u of dy
of course I did it for the gradient I could have done it for any first order partial derivative
but actually I can do it for any order derivative because whenever you prove it for first order
derivatives if you know that small f here is such that the derivative over there alpha for any
multi-index alpha exists for almost every x u and all v y and v the the derivative of order alpha
minus ej for some j is l1 in some point and you have this bound for the alpha derivative then
all of this works the same basically you just take here the function integration of the
derivative over there alpha minus ej and then you take another derivative if you apply this
if you want some more explanation some if you need some to see the proof of these results please
let me know of course I can spend some more time on it that I was giving these things kind of
granted and I should have put it in the first slides I agree on this sorry so let us come
finally to the first exercise n is larger than 2 but also the second exercise rn plus is the
half space x where xn is positive its boundary is the hyperplane xn equal to 0 which can be
identified with rn minus 1 times 0 we define the Poisson kernel in rn plus to be k of x y
defined as 2 xn over n omega n times 1 over norm of x minus y to the power n
for g in l infinity of the boundary of rn plus we set u of x to be the integral over the boundary
of rn plus of k x y g of y integrated with respect to the surface measure
yes then you have to prove that the integral over the boundary of rn plus of k x y in the s of y
is equal to 1 for all x in rn plus u is in l infinity of rn plus with this bound u is actually c
infinity of rn plus and oh I'm sorry here there is the old notation this should be the alpha
sorry this is the notation of harmonic analysis okay this is the partial derivative the alpha
then I wrote it correctly and the Laplacian of u of x is equal to the integral over the
Laplacian of k but the Laplacian of k is 0 so u is harmonic finally this was optional take
a function g which is not continuous because notice I asked only l infinity of the boundary
not continuous and show that g of y see if you take this g of y which is 1 if yn minus 1 is
positive and 0 otherwise and there's also 1 if yn minus 1 is 0 but doesn't matter then you should
find the limit of the function ux1 x2 xn as xn goes to 0 plus so you see this function is
discontinuous so you cannot say that u of x converges to g of x1 x2 xn minus 1 as xn goes to 0 plus
you need to be careful just before coming to the exercise a little moment of reflection on what
does it mean l infinity of the boundary of rn plus we already explained this during the online
week limiting but given that not many of you come to listen it's better to repeat ourselves
so this space should be rigorously defined as the space of measurable functions from
the boundary of rn plus in 2r which are essentially bounded with respect to a surface measure
the house of n minus 1 dimensional measure restricted to the surface to the boundary of
rn plus which is basically the layback measure this is the formal definition but because the
boundary of rn plus is equal to rn minus 1 times 0 basically g is in the infinity of a boundary of
rn plus if and only if the existence function h in l infinity of rn minus 1 with respect to the
ln minus 1 the back measure such that g of z 0 is equal to h of z follow z in rn minus 1
so often this identification is done immediately especially this happens in evan's book of partial
differential equations i just wanted to be a little more rigorous for you the same argument
by the way works with the surface of any open set with regular enough boundary the ball for instance
or i mean any Lipschitz boundary is fine and then one could extend all this argument and repeat it
the case p less than infinity by taking the as usual the integral over the boundary of rn plus
of the function to the power p with respect to the surface measure but this is equal to taking
the integral on rn minus 1 of gz 0 with respect to the power p with respect to some layback measure