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It is well known (also known as Schauder's Theorem) that if $X$ and $Y$ are normed spaces and $T:X\to Y$ is a linear and compact operator, then also $T^*:Y^*\to X^*$ is compact. The converse is true if $Y$ is complete.

So the natural question is:

Is there an "easy" example that shows that we cannot drop the completeness of $Y$ for the converse implication?

In order to prove the converse, one usually applies the first implication to the bidual. Thus, a counterexample should be rooted in the subtle difference between "relatively compact" and "totally bounded", but I cannot wrap my head around it.

Any ideas are highly appreciated. Thank you in advance!

Bernard
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sranthrop
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    I don't think you need completeness. Isn't $T$ the restriction to $X$ of $T^{}:X^{} \to Y^{**}$? – Kavi Rama Murthy May 24 '19 at 09:32
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    If $T:X\rightarrow Y$ is a compact operator such that $T(B_X)$ is not closed and $T(X)$ is dense in $Y$, a counterexample can be made. Let $S$ be $T$ regarded as an operator into its range. $S$ won't be compact, but its adjoint will be the same as $T^*$. – David Mitra May 24 '19 at 18:09

2 Answers2

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Edit: This does not work.

Here is a proof using completeness of $Y$: We have that $T^{**} : X^{**} \to Y^{**}$ is compact. Thus, $i_Y \, T = T^{**} \, i_X$ is compact (where $i_Y : Y \to Y^{**}$ and $i_X : Y \to X^{**}$ are the canonical embeddings). Hence, if $(x_n)$ is a bounded sequence, $(T^{**} \, i_X \, x_n)$ has a convergent subsequence indexed by $(n_k)$. Since $i_Y$ is an isometry, $T \, x_{n_k}$ is Cauchy in $Y$.

gerw
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  • According to the comments in https://math.stackexchange.com/questions/2357258/is-my-proof-for-schauders-theorem-in-non-banach-spaces-correct?rq=1, this idea cannot work. – sranthrop May 24 '19 at 13:28
  • If $Y$ isn't complete, then neither is $i_Y$. I can see only that $(T^{**}i_X x_n)$ has a limit point in the closure of $ i_Y (Y)$. – David Mitra May 24 '19 at 13:29
  • We only use that $i_Y$ is an isometry. Suppose that the entire sequence $T^{} i_X x_n$ converges in $Y^{}$. Then, it is a Cauchy sequence in $Y^{}$. Now, since $T^{} i_x = i_Y T$, also $i_Y T x_n$ is a Cauchy sequence in $Y^{**}$. Since $i_Y$ is an isometry, $T x_n$ is Cauchy in $Y$. Ok, and here enters completeness! – gerw May 24 '19 at 15:53
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Let $X=c_0$, $(Tx)_k=x_k/k$, and $Y=\mathrm{im}\,T\subset\ell^2$. Then $T$ is the norm limit of finite-rank operators, so $T^*:Y^*\to X^*$ is the norm limit of compact operators, thus compact (as $X^*$ is complete).

Let $x_k^{(n)}=1_{k\le n}$, then $\|x^{(n)}\|_{c_0}=1$ and $Tx^{(n)}\to y\in\ell^2\setminus Y$, where $y_k=1/k$. So $(Tx^{(n)})$ has no convergent subsequences. Thus $T$ is not compact.

Yizhen Chen
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