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Schauder's theorem states:

Let $X, Y$ be Banach spaces, let $T \in B(X, Y)$ be a bounded linear operator.

Then $T$ is compact $\iff$ $T'$ is compact, where $T' \in B(Y', X')$ is the dual operator.

The $\implies$ direction is true if $X, Y$ are just normed spaces, see for example E. Kreyszig, Introductory Functional Analysis with Applications, Theorem 8.2-5, pp. 416 (you can find it on google).

For the other direction I have a proof that also seems to work if $X, Y$ are just normed spaces. Is it correct?

Let $T' \in B(Y', X')$ be compact. Using "$\implies$" we have that $T'': \in B(X'', Y'')$ is compact. Let $J_X \colon X \to X''$ be the canonical embedding, $J_Y$ similarly. It is known that $$ T'' J_X = J_Y T $$ It follows that $T = J_Y^{-1} T'' J_X$, which is well defined if you consider $J_Y^{-1} : J_y(Y) \to Y$ which is linear and bounded since it's norm-preserving. But then $T$ is compact since $T''$ is compact (If $A$ is compact and $B$ and $C$ are bounded operators, then $BAC$ is compact).

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  • $\begingroup$ why does compactness of $T''$ imply compactness of $T$? $\endgroup$ Jul 13, 2017 at 11:27
  • $\begingroup$ @uniquesolution If $A$ is compact and $B$ and $C$ are bounded operators, then $BAC$ is compact, see en.wikipedia.org/wiki/Compact_operator $\endgroup$
    – Vincent
    Jul 13, 2017 at 11:34
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    $\begingroup$ Strictly speaking you are not in this situation, since $J_Y^{-1}$ is not a map from $Y''\to Y$. However this is not a problem. Another way to do it is to look at the image of the unit ball in $X$ under $T$, that is $T(B_1(0)_X)$, this is a subset of $T''(B_1(0)_{X''})$ when you view $Y$ as a subspace of $Y''$. Since $T''(B_1(0)_{X''})$ is relatively compact then any subset of it is relatively compact, so $T$ is compact. $\endgroup$
    – s.harp
    Jul 13, 2017 at 11:49
  • $\begingroup$ @s.harp Thank you very much! Though I do believe this is a problem. By viewing $T(B_1(0))_X$ as a subspace of $T''(B_1(0))_X''$, you are essentially claiming that J_Y preserves the relative compactness. I was unable to justify that without using that $J_Y(Y)$ is closed, i.e. $Y$ is a banach space. $\endgroup$
    – Vincent
    Jul 13, 2017 at 12:34
  • $\begingroup$ Oh, somehow I missed that you were interested in the non-Banach space situation. Indeed it does not need to be true, look at $X=c_{00}(\Bbb N)$ with norm $\|x\|=\sum_n 2^{n}|x_n|$ and $Y=c_{00}(\Bbb N)$ with norm $\|x\|=\sum_n |x|$. The identity is not a compact operator, but if you look at the completions of $X$ and $Y$ I think the extension of the identity is compact, so the adjoint operator (which is the same whether you start with the extension or the original) is compact. This should be a counter example. $\endgroup$
    – s.harp
    Jul 13, 2017 at 15:32

1 Answer 1

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As s.harp pointed out in the comments, the problem is in the last sentence, when claiming that $T = J_Y^{-1} T'' J_X$ is compact because $T''$ is.

The theorem goes:

If $K \in B(X, Y)$ is compact, $H \in B(W, X)$, $S \in B(Y, Z)$, then $SKH \in B(W, Z)$ is compact.

Because of this, $T'' J_X$ is clearly compact. But because $J_Y^{-1}$ is defined on $J_Y(Y)$, if we want to conclude that $T$ is compact, we must have that $T'' J_X$ is compact even if we view it as a function from $X$ to $J_Y(Y)$. But it is not obvious that it is still compact then, and probably not true in general. What we can say is that it all works out if $J_Y(Y)$ is closed in $Y$: If $x_n$ is a bounded sequence in $X$, $T''J_X(x_n)$ has a convergent subsequence because the "regular" $T''J_X$ is compact. Since $J_Y(Y)$ is closed, the subsequence is convergent even in $J_Y(Y)$.

An equivalent condition to $J_Y(Y)$ being closed is $Y$ being a banach space.

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