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!

• I don't think you need completeness. Isn't $T$ the restriction to $X$ of $T^{**}:X^{**} \to Y^{**}$? Commented May 24, 2019 at 9:32
• 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 $T^*$. Commented May 24, 2019 at 18:09

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 $$Y$$.

• According to the comments in math.stackexchange.com/questions/2357258/…, this idea cannot work. Commented May 24, 2019 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 $i_Y (Y)$. Commented May 24, 2019 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
Commented May 24, 2019 at 15:53

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.