Converse of Schauder's Theorem about compactness of adjoint operator

Question

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!

Answer 1

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 $Y$.

Answer 2

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.