Dear students, hello. In this exercise class we shall see, as usually, the solution to
last exercise sheet, and this time it was all about the application of the theory of
sovereign spaces to actually proving the existence of weak solutions of some type of partial
differential equations, I mean those of elliptic type of course, in very specific sovereign
space which means H2 and H1. So these are the general definitions and we use again the
notation du and d squared u and Laplacian u to denote, respectively, the weak gradient
and the weak Hessian matrix and the weak Laplacian. I mean, in this case, no. Ambiguity, there
is no ambiguity. For, in order to solve true of the exercise, it was important to use Poincare
inequalities in two particular cases, in the case of the zero boundary values in which
we assume open set u to be only bounded, and then we have the Drexis, the constant c depending
on n, p and u such that the lp norm of small u over capital U is controlled by this constant
times the lp norm of the gradient of u for all u and w1, p0, and instead the Poincare
inequality in the zero mean value case, that means if we have u which is open bounded connected
with c1 boundary, then the lp norm of the function u and w1, p minus its average, which
is part of the measure of the domain times the integral of u over the domain, is controlled
by the constant times the lp norm of the u. Notice, just curiosity, but why do we need
such a regularity of u here and of the domain u here, and here we don't? But roughly speaking,
because here you has zero boundary values, you know, in the weak sense, in the sense
that it is, it can be seen as the limit of c infinity function with respect to w1, p norm
in this sense. Note that it has pointwise values well defined and in the classical sense.
It's a trace, yes, but it's a little softer than that, and so in a sense the boundary view doesn't
play a role in this inequality. Instead, here we don't tell such assumptions, so the boundary is
important. Also, one can ask, well, okay, but then the connectedness, of course the connectedness
is not related to the boundary, but the point is if we don't put this assumption,
this statement is clearly false, because take as an open set, which is, we see my boundary, I'm
bounded, but not connected. So, for instance, choose these joint balls and take u to be constant,
equal to one and to, I don't know, any other number, a different one on the other board, zero,
if you want, and then the average of u will be,
of course, it will be one out, I mean, no, it depends if the two balls are the same radius,
it will be one over the measure of the two balls, the sum of the measure of the balls,
times the measure of the ball in which u is one, so it will be a number in zero one, so here they
will be written the integral of a function which is either one or zero minus lambda in zero one,
so this is a non-zero integral, while here, while the gradient of this function will be zero, of
course, that this doesn't make sense. I mean, if you don't know if you already saw the proof of
this statement, but basically it somehow, it requires that this, it's a proof by contradiction,
which requires the fact that in a way you must find a contradiction working on the fact that
you would have a sequence of functions which on the one hand converges to
a non-zero function, on the other hand it converges to something with zero gradient,
so the function must be constant by constant with zero average means zero. The trick is that,
but this was not relevant to the solution of the exercise, it's really important to your knowledge.
So I will set our work, especially in the case p equal to two, and in the case we denote hk of u to
be wk2, and hk zero to be wk2 zero. These are Hilbert spaces, once equipped with a scalar
product given by the sum of the scalar products in a true between any alpha weak derivative for
any multi-index alpha with the length less equal than k, and we denote by h minus one of u the dual
space up to h1 zero, and we write in this way the pairing between the element of h minus one of u
and h1 zero. That's important because h minus one of u, seeing as the dual of the, of course,
is a Hilbert space, so it's isometric to h1 zero, of course, because all Hilbert spaces are isometric
to each other, all separable Hilbert spaces, sorry. But if you want to see it as the dual of h1 zero
and not to represent it in a way that it is equivalent to it, you get a rather different
way to write elements of h1 zero, and you saw this in the lecture with Professor Nobili.
So the first exercise consists in showing that if we take an element v of l2, then the pairing in,
with respect to in h minus one between v and an element u of h1 zero is equal to the l2,
l2 scalar product between v and u, and this implies that l2 of u is inside h minus one of u. So there
were many ways of doing it. One could follow what it was done in the lecture, which is basically
using the results from Evans' book, or here I have two other ideas. Thanks to this representation
theorem, one could say that, look, the dual of l2, which is l2 star, is actually l2.
This means, you know, this function arises, any linear continuous function on l2 can be
represented by the scalar product against some fixed element in l2. And so the inclusion h1 zero
inside l2, which is obvious, immediately else, you know, when you take the dual, you reverse
the chain of inclusions between the inverter, actually, dynamic spaces, and so you get limited
and true must be inside h minus one. And so that any element in l2 actually defines a scalar
product in h minus one. On the other hand, you can also notice that if v is in l2, let's consider,
you know, you have to prove this equality. So let's consider this type of function,
the function that sends each u in h1 zero in the scalar product against this v. This, of course,
is linear continuous, but not only on l2 also in h1 zero, because, well, of course, it's what's
inequality, we get l2 norm of u, v, sorry, l2 norm of u, but the l2 norm of u is controlled by the
h1 zero, h1, and the equivalent norm of u, trigaly. And this, therefore, the scalar product against
this v in respect to the l2 Gilbert structure is an element of h minus one, because integration,
sorry, because integration against an element of h1 zero gives a linear continuous function on h1
zero, which is the definition of h minus one. This proves the inclusion, each element v in l2
gives an actual element in h minus one, and the identity, because we denote in this way the
pairing between elements of h minus one and elements of h1 zero, and so on.