Finally, the last exercise concerned the approximation of LP functions by smooth
modification. We said that the function eta is a standard
modifier if it is in C infinity C, so smooth with complex support of the unit ball is rather
symmetric if C integral is equal to 1. For epsilon positive we set e to epsilon to be
epsilon to the minus n eta over x over epsilon, so the rescaled version of eta. Given a function
of one lock, we define the magnification of epsilon simply as the convolution with eta
epsilon, which of course makes sense as long as you stay far from the boundary of u, the
domain u, much far, the distance should be larger than epsilon. And then your exercise
was to prove that if p is in one infinity and f is in LP lock, then f epsilon converges
to f in LP block of u, which means a strong LP convergence in every open subset of u which
is compactly contained in u, and therefore let u and v be sets in u open such that v
is compactly contained in w, compactly contained in u. Why do you need w here, you may wonder.
The key point is that you want to be far enough from the boundary of u, so let's say you
can always be as far as be able to put another open set in between. The starting trick is
check that the LP norm of the magnification is controlled by the LP norm of a function
up to a constant and up to getting a slightly larger domain. So if we take epsilon such
that the union of all both centered in points of v of radius epsilon is inside w, which
means the union of the set of points was distance from v is less than epsilon is contained
in w. This is always possible because there's a relation. Then the LP norm of f epsilon
over v will be less equal than a constant, omega n is the dimension of the, sorry, the
volume of the unit ball, del infinity norm of eta times the LP norm of f over the slightly
larger domain w. How to prove this? Fix x in v and estimate the absolute value of f
epsilon of x in this way. Bring the absolute value inside, write eta epsilon in this way,
use Helder inequality here so that we get the absolute value of f to a power p, which
is what we like because then we need to estimate the LP norm. Now, in some cases eta is assumed
to be non-negative. For instance, in the proof of this result in Evans book is like this.
In that case, it's easier because you say, ah, but the integral of eta is equal to one and so for
sure this is one and we will not give so much troubles. Sure. However, I notice it works also
if eta is as a sign because it is bounded. Of course, eta epsilon is one over epsilon to the n,
eta rescaled, but this doesn't change the LP norm of eta. There is a factor one over epsilon to the
n of course, but the measure of the ball B x epsilon is omega n epsilon to the n. So epsilon to the
minus n and plus n simplify. So you get with this estimate the measure of the unit ball to the power
one minus one over p times the LP norm of eta to the power one minus one over p. But then you
estimate also this is the same way as you get power one. And here you have a factor one over epsilon
to the n which remains into play. Now, take the integral over V of the absolute value of epsilon
of x to the power p, the x, this will be a sequel thanks to the estimate by, well, this constant to
power p, one over epsilon to the n integral over V integral over B x epsilon of the absolute value
of f to the power p in the y dx. And EDA is exchanged the order of some integration.
So for y, for x fixed, the y is moving in B x epsilon. Therefore, the set in which y live
for sure is contained in W because of this condition over here. Therefore, this can be
estimated from above by the integral constant one over epsilon to the n integral over W of
absolute value of f to the power p times integral over B y epsilon of one in the x.
This gives me omega n epsilon to the n and simplifies epsilon to the n and EDA gives
omega n to the power p. But most importantly, this is the Lp norm of f to the power p over W.
Now, take one over p and we are done. To conclude, we need to use the density of the
continuous functions up to the boundary of W and Lp, which means if I draw delta positive, I can
find g continuous up to the boundary such that the Lp norm of f minus g is strictly less than delta
and this is why we need p to be strictly less than infinity because otherwise it's false.
And now the trick is sum and subtract. So we have to estimate the Lp norm of f epsilon minus f
sum and subtract g epsilon and g. So we estimate this by the Lp norm of f epsilon minus g epsilon
plus g epsilon minus g plus g minus f. But this is, by the linearity of the convolution product,
is simply the Lp norm of f minus g epsilon. And we know that the Lp norm of the convolution of
the modification can be controlled by a constant times the Lp norm of the function over W,
the slightly larger set. Of course, the Lp norm of f minus g can be controlled by the Lp
norm of f minus g over W because W is larger than v. So we get this constant times this Lp norm
and then g epsilon minus g epsilon over v. But this is controlled by delta
assumption. So this is arbitrarily small. So that if we take, now remember that
since g is continuous, then g epsilon over W, not the closure zone, then g epsilon is converging to
g uniformly on compact subset of W and v is compactly contained in W. It means its closure
is a compact subset of W. And therefore, this last term is going to zero by uniform convergence.
Therefore, if I take the lim sup, I get that the lim sup as epsilon goes to zero,
the Lp norm of f epsilon minus f norm of v is controlled by a constant, which, I mean,
it is fixed and so on. This is all a finite number times delta. But delta is arbitrary,
so we can set it to zero. And this gives our conclusion. Phil and Dank for the Aufner sum guide.