WEBVTT - autoGenerated
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Dear students, hello. Today we are going to see some sufficient condition which allows
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us to ensure the existence of the minimizer of functionals we are going to consider, that
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is the integral functionals we studied in the previous lectures. This sufficient condition
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which ensures in particular the boundedness and therefore in a suitable topology a type
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of compactness is the notion of coercivity. Given a normed space X and a subset A of X
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and the functional F from A into the extended real line, then we say that F is coercive
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in A or better X coercive, because it's coercive with respect to the norm on X, if the limit
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of F of W as the norm of F goes to plus infinity for W in A is plus infinity. This means that
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for every epsilon positive there exists some delta positive such that for every W in A
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with the norm larger than one over delta, then the value of F of W must be larger than
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one over epsilon, the definition of limits. So what does it mean to take a limit in this
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more abstract context? So this we're not going to prove because it will be an exercise, but
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one can prove that the following statements are equivalent, that is, F is X-squaresive. Given
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any sequence WK such that the supremum of F of WK is finite, then the sequence itself must be
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bounded, uniformly bounded. Every sublevel set of the functional F, that is a set of W in A such
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that the F of W is a secant and S for some S in R is a bounded set with respect to the norm of
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course. Now, remember in the general existence T or M, we were given among the conditions. So one
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of the conditions was the lower semi-continuity topological or sequential, and the other condition
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again topological or sequential was the compactness or the compactness of the sublevel set or the
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sequential compactness. And what do we have? That coercivity gives us for free boundedness of the
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sublevel sets of F. Actually, it's an if and only if, so it's even in a sense showing that the
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coercivity is in some sense necessary. And therefore, this is the point in which the choice
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of a weak topology plays a crucial role because of Banach law of Routhier. That is, if our space
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X is a reflexive Banach space, then every closed ball is weakly compact. And if X is the dual of a
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normed space, then every closed ball is weakly star compact. In other words, in a reflexive space,
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all bounded sets are weakly relatively compact because they can stay inside the ball, the ball,
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the closed ball, they can stay inside the closed ball, the closed ball is compact. And so with
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respect to this topology, the set must be weakly compact. I mean, the closure of the set must be
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weakly compact. And in addition, in the dual of a normed space, all bounded sets are weakly star
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relatively compact. And therefore, coercivity, even though this notion depends only on the norm,
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see the definition, it depends only on the choice of a normal X, not on the weak topologies,
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nevertheless, gives the required weak compactness under these two additional assumptions that is
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the fact that X is a Banach space, so the completeness and the reflexivity, or that X is the dual of a
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normed space. One can argue which one of the two is the nicest, it depends on the cases. So this
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would be a quite general setting in which we have what we wanted to achieve with our initial
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existence result. Of course, this statement can be given again in a sequential version,
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let's just mention it for completeness. Again, a weak and a weak star version, every closed ball
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in a reflexive Banach space is sequentially weakly compact. What does it mean? That if I find a
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bounded sequence in a reflexive Banach space, then it must have a weakly convergent subsequence. And
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weak star version, every closed ball in a dual of a separable normed space, separable is important,
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is sequentially weakly star compact. And this means every bounded sequence in a dual of a
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separable normed space must have a weakly star convergent subsequence. Why do we need separability
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in this second case? Because there is a simple counterexample. So consider the dual of a small
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L infinity. Small L infinity is not separable, you cannot find a dense countable subset. This is
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it's a simple exercise from functional analysis. We define this family of functionals EK on L infinity
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in this simple way, they are the evaluation function. So EK applied to some sequence
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xj in L infinity give us the kth element of the sequence. EKs are clearly linear continuous
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functionals, they're actually bounded and the norm is of course one, so they belong to the
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unit ball of L infinity star. But, and this is the important point, given any subsequence EKL,
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the limit as L goes to plus infinity of EKL applied to some elements of L infinity, so the limit in
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the sense of weakly star compactness, weakly star topology, does not exist. That is, we cannot,
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they don't exist in general. That is, we cannot find a subsequence of EK which for sure is weakly
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star convergent. Because given any subsequence EKL, I can find a sequence xj in L infinity such
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that the limit of EKL applied to xj does not exist. For instance, choose xj in L infinity such that xkl
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is minus one to the L. Of course this is a suitable subsequence, I don't know for instance,
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for instance you can set all other values of j equal, xj equal to zero, because it's
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bound, so it's of course in L infinity, it's bounded and so we're done.
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Now let's come to what we want to achieve, that is weak compactness. Consider therefore a reflexive
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Banach space X is of set A in X and a functional F from A into the extended real line. So if A is
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weakly closed in X, then the following are equivalent, TFAE is the shortening for this
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expression. F is X square given A is equivalent to every sublevel set of F is weakly relatively
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compact in A. And true, if A is sequentially weakly closed in X, then the coercivity of F
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is equivalent to the fact that every sequence in A such that the supremum of F WK is finite
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as a weakly convergent subsequence in A. In addition, this is the obvious second case, if
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we consider X star, so A is a subset of a dual of a separable Banach space X, then
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the X star coercivity of the functional F is equivalent to the synchronization you give above
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replacing X star with X. We should prove only the first two statements, the second being analogous.
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I mean the case for dual of a separable Banach space being analogous at least in the idea of the
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proof. So for this proof, it is essential to exploit another very important tool from
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functional analysis, which is the uniform boundness principle. The general uniform boundness principle
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says that if I have a Banach space X, a normed space Y, and a family of continuous linear
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functionals, I mean, sorry, operators, sorry, continuous linear operators from X into Y,
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this is the symbol L, X, Y, such that the pointwise supremum is finite, which means the supremum of
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all operators T in this family F of the Y norm of TX is finite for every fixed X in X. So the
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pointwise supremum with respect to a norm of Y of this family is finite implies that this family is
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bounded with respect to the operatorial norm in L, X, Y. How does this apply in this case? Well,
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if I take X to be a Banach space and S to be a subset of the dual, which is weakly star bounded
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in a sense that the supremum of all element X star in S, which here they work as, of course,
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as operators, linear continuous operator from X into R. So if the supremum of X star in S of the
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absolute value of X star applied to X is finite for all X in X, this condition here clearly,
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then we get that the supremum over X star in S of the norm of X star in the dual is finite,
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but this is exactly, means exactly that S is a bounded set with respect to the dual norm.
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In addition, yet again another consequence, if X is a normed space and S in X is weakly
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bounded, what does it mean weakly bounded? It means that the supremum over all X in S of the
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absolute value of X star applied to X is finite for all X star in the dual. So weakly bounded means
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whenever I consider the image of this set through a continuous linear functional, then this image is
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bounded, then the image J X S under the canonical isometric embeddings, this is the
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the B-dual map, the map which isometrically embeds X into the B-dual X star star,
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must be weakly star bounded in X star star. This is clearly over here we get the
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the notion of weakly star boundedness, because each element X in S through this map becomes an
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element of X star star, but then if we apply what we just saw above, we're replacing X star,
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putting X star instead of X. This gives us the fact that J X of S is bounded with respect to the
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norm in the B-dual space X star star, but thanks to the isometrical embedding this means that S
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must be bounded with respect to the norm of X itself. I hope this is clear, but take your
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time to understand this. Because it's fundamental now, proof of the corollary, coercivity is
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characterized, in a sense it's equivalent to boundedness of the sublevel set of the functional,
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hence since A is weakly closed, remember, it's enough to show that, to prove statement,
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first statement, it's enough to show that a set in A is bounded if and only if it is weakly
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relatively compact, because the sublevel sets are bounded thanks to coercivity, so
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we can just rely on these equivalence. And why so? Forward implication, so S is bounded,
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therefore S is inside a suitably large closed ball in X. Perfect. Banachalogrutirim, remember
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that X is reflexive, therefore B is weakly compact, but then the weak closure of S inside B
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must be weakly closed, I mean, obviously, and it's still contained in B. But therefore it is weakly
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compact. This is a general statement, if you have a closed set inside a compact set with respect to
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the given topology it must be compact with respect to that topology. Therefore, since A is weakly
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closed, the weak closure of S in A is equal to the closure of S, the weak closure of S in X.
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In other words, it does not get out of A. S is a subset of A, which is weakly closed, therefore
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the weak closure of F is inside A still, and therefore S must be weakly relatively compact in A.
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And we are done, because the fact that the weak closure of S is weakly relatively compact,
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we derived one line before. Backward implication, so assume that S is weakly relatively compact.
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What does it mean? It means that this closure is weakly compact in A. There,
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fix x star in the dual of x. Of course, this is a function, so is weakly continuous. The image,
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I mean, it's an element of the dual, so it's a linear continuous function,
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continuous in the sense of weakly continuous. Therefore, the image of S with respect to x star,
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that is the set of real number x star of x for x in S, must be relatively compact in R.
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Now, a continuous map between two topologies, maps, relatively compact set into compact set,
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and therefore relatively compact set into relatively compact set. One of the equivalent
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definitions of continuity, again topology. And thus, what we get is that the supremum
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of x in S of the absolute value of x star applied to x must be finite for all x star in X.
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Because we just said that fixed this element x star of the dual, the image x star of S
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is a relatively compact set in R. But what is a relatively compact set in R is a set whose
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closure is compact. But the compact set in R is bounded. If the closure of a set is bounded,
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also the set itself must be bounded. And so, this is a way to rephrase this boundedness.
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But this allows to apply the uniform boundedness principle. In other words, we get that S is weakly
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bounded, and so, thanks to uniform boundedness principle, S must be weakly, I mean, sorry,
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not only weakly, it must be bounded with respect to the norm of the space x itself.
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And so, we're done. Let us scan now to statement 2. Again, here it is enough to prove that
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the set S in A is bounded if and only if every sequence in S has a weakly convergence of sequence
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in A, because again, S does not need to be closed. Indeed, as we saw,
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coercitivity implies boundedness of the level set, and the condition that the sequence WK satisfies
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the supremum over K in N of F of WK to be finite implies that WK is in a sublevel set.
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Because remember what we need to show. We need to show that given a sequence WK in A,
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such that the supremum over K in N of F of WK is finite, must have a weakly convergent
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subsequence in A with limit in A. Therefore, F x-squares means sublevel set of F are
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bounded. But then, if I take a sequence with this property, it must belong, for sure,
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to a sublevel set of F and therefore to a bounded set. And therefore, it is enough to prove that
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every bounded set in A is such that every sequence in this set has a weakly convergent
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subsequence with limit in A. So, forward implication. Notice that S is bounded.
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So, we take a sequence in this bounded set, and this must have a weakly convergent subsequence.
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Why? By the sequential Banach-Hannog-Luthier. We just saw. And therefore, there exists WKL,
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say, such that it has a limit W. Where does this W belong? Well, to A, because A is sequentially
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weakly closed by assumption. And so, we are done. Opposite implication. So, assume that every sequence
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in S has a weakly convergent subsequence. Now, we need to show that S is indeed bounded.
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But now, this is done again by the uniform boundedness principle. Because the uniform boundedness
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principle implies that the weakly convergent subsequences must be bounded. So that every
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sequence in S has a bounded subsequence. Now, because we assume that every sequence in S has
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a weakly convergent subsequence. The uniform boundedness principle tells us that every
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weakly convergent sequence must be bounded, and therefore, every sequence in S has a bounded
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subsequence in particular. But this is another way to express the fact that S is bounded,
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because what does it mean that S is unbounded? It means that there exists a sequence of elements,
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WK in S, such that the limit of the normal WK, sorry, here is a mistake, is plus infinity.
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But, we just said that given any sequence in S, it must have a bounded subsequence.
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Now, assume that there exists a sequence WK, such that the limit of the norm exists,
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and it is plus infinity. And you see immediately that here there is a contradiction, because if
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not, you will be able to get here a subsequence WKL, such that the norm of WKL stays bounded.
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And but this is in contradiction with the fact that the full limit of the sequence is plus infinity.
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And then finally, the weak start versions can be obtained, and obviously,
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they are left to you if you want.