WEBVTT - autoGenerated
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Exercise 3 dealt with Poincare inequality in the zero boundary value case
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and in particular you had to prove not only the Poincare inequality but actually
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find a precise formula for the constant
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which not necessarily gives you a sharp constant in all the cases
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but it allows to get a rather general result
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first some definitions. We say that set E in Rn
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lies in a strip of finite width if there exists a unit vector nu
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and some numbers in R a and b with a less than b such that for Laxine E
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the scalar product between nu and x is between a and b
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or E is inside the strip S a b
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which is defined as the set of point x in Rn such that again this condition
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what does it mean this condition? It means that the point x
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is in between the hyperplane nu dot x equal b
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and nu dot x equal a simply
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and how is it defined the width of the strip of course is
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the difference between b and a and the orientation is given by nu
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now let Omega be an open set which lies in a strip of finite width
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and p in one infinity you have to prove Poincare inequality that is
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for all u in w and p zero of Omega the LP norm of u is controlled by a constant
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c and p Omega such that times the LP norm of the u
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where c and p Omega is defined as p to the power
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minus one over p times the infimum of the width
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of the strip S a b nu such that Omega is inside S a b nu
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yeah here I didn't put a less than b but it's obvious maybe
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the definition of the strip should have this condition
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very well so before you start to do this proof in a
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full generality notice that it's possible to simplify
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a little the setting first of all thanks to the density of c
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infinity c in w1p0 by definition because w1p0 is the closure
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of c infinity c with respect to the w1p norm therefore
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we know that given u in w1p0 there exists a sequence uj
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in c infinity c such that u minus uj converges to 0
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in w1p therefore this means that we can work simply with v in c infinity
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why because if we are able to prove that for every v in c infinity c we have the
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Poincare inequality with that constant then given u in w1p0
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the lp norm of u can be controlled trivially by the
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lp norm of u minus uj plus the lp norm of uj where uj is the approximate
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sequence in infinity c but then the lp norm of uj can be bounded
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in this way with the constant time the lp norm of d uj
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now we send j to plus infinity the lp norm of u minus uj goes to 0
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while the lp norm of du the uj goes to the lp norm of du
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because uj and duj are converging in lp to u and du
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and so if we're able to prove the inequality for c infinity c function
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thanks to this simple argument we are done therefore for now on we assume
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u to be smooth with compact support in omega
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in addition we know that since omega is a set of finite width
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then there exists at least one strip with finite width
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sa v nu such that omega is inside sa v nu
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let r be a rotation such a rotation matrix of course such that
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r nu is equal to e1 the first canonical vector so 1 0 0 0
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then we have that r applied to omega so the rotation of omega
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is inside the rotation of the strip sa v nu but how is it written this
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it's the set of y which is equal to rx such that x is inside a sa v nu so
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x is in Rn and nu dot x is between a and b
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but since x since y is rx x is r transpose y
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but the roots of scalar product nu dot matrix y is equal to
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transpose of the matrix applied to nu against y
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scalar product against y but by definition r nu is equal to e1
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so we see that the rotation of sa v nu is simply
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sa v e1 so the set of y in Rn such that y1 is between a and b
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and why all of this because that's much simpler so the set of
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points in Rn such that the first variable is between first coordinate is
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between a and b for some a and b fixed in R
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it's much simpler to work in this setting in addition notice that if u is
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move with complex support c1 of course was enough
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and we define ur of x to be u of r transpose x
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then ur is of course in infinity c of r omega
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and the gradient is equal to the transpose of a gradient
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against the gradient of u evaluated in Rx
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since of course the lp norm of u and the lp norm of ur and the lp
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norm of d ur are the same are equal to the lp norm of u and the
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lp norm of du because i mean the rotation if you apply
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again the rotation the rotation is a matrix with the determinant
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equal to one and so it doesn't change anything
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we can assume without loss of generality that
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and also it's an isometry we can assume without loss of generality that
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omega is already inside the strip sab e1 so the set of points in Rn
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such that x1 is between a and b this all of this simplifies a lot of work
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because now we have u in infinity c of omega which therefore is in infinity c
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of sab e1 and we decide to use the notation
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these notation points x are written as x1 x prime
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where x prime is in Rn minus one since u is in this
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is with the compact support inside the strip it means that ua
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x prime is equal to zero for every x prime in Rn minus one
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of course it's zero near by the boundary therefore u of x
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is equal to u x1 x prime minus u of a x prime is equal to zero so it's fine
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and therefore because of smoothness the fundamental theorem of calculus
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yields this is equal to the integral from a to x1
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of the first component of the gradient or
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e1 dot du i mean notation every way the dt x prime dt
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and this works for every x1 in ab and x prime in Rn minus one so that
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x is in the strip therefore take the absolute value take the power p
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p is fine we can do it the absolute value of u
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of x to the power p is controlled by the integral from a to x1 of
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bring inside the absolute value exaggerate put to the gradient
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of t x prime in the t to the power p now use
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herder inequality and so when we get the integral from a to x1
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the norm of du to the power p t x prime
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it will be all to the power one over p but there is p here so it simplifies
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and then times the integral of one gives x1 minus a to the power one minus
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one over p again raised to the power p so it becomes
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p minus one now integrate everything over Rn minus one
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so x1 now is fixed so this term is not touched by this integration
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and now we have an integration over Rn minus one over of the integration from a
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to x1 of norm of du to the power p t x prime dt
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dx prime now exaggerate instead of x because x1 at
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most is b so put b and this is actually the integration
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over the strip s ab e1 of du of the norm of du to the power p
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but remember u is infinity c in omega so this is actually integral over omega
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of the gradient of u finally integrate on the left hand side also over ab in x1
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because the integral over s ab e1 of the not absolute value of u to the power p
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is equal to the integral from a to b of the integral over Rn minus one of
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the absolute value of u of x1 x prime to the power p in dx prime dx1
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and this is controlled by the integral over ab of x1 minus a to the power p
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minus one in dx1 times the integral over the full strip of
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the norm of the u to the power p and this first term of course give us b
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minus a to the power p divided by p therefore take the power one over p of
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both sides because we need to take the lp norm we have that the lp norm of u
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over omega is equal to the lp norm of u over the strip
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s ab e1 which is controlled by b minus a over p
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to the power one over p and now times the lp norm of du
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with over the strip which is of course equal to lp norm of u
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of du over omega and now b minus a is the width of the strip
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and now we are done we get this inequality
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for u is infinity c of omega and for any a b in r
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with a less than b such that x t is new in Rn
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with absolute value of nu equal to one such that omega is inside the strip
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s ab nu but therefore i mean we selected one of these strips of course
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there can be there can be only one in case for instance
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omega is a strip omega is the set this set
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but omega could be a bounded set so that there are
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infinitely many strips of finite width that contains omega
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and so since this argument can be clearly done in any one of these
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directions we can take the infimum so in other
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sense the real optimal constant under this
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point of view will be given by the smallest
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the width the smallest possible width of omega
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with respect to all possible directions and we are done