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