I'm trying the prove that if $X$ is a normed space and $Y$ a Banach space then a linear operator $A:X \rightarrow Y$ is compact if and only if the image of the unit sphere is sequentially compact.
My efforts so far:
Suppose $A$ is compact. Let $(y_n)$ be a sequence in $A(S_X)$, the image of the unit sphere $S_X$ in $X$. Then for each $n$ there exists an $x_n \in S_X$ such that $y_n = A(s_n)$. Then $(x_n)$ is a bounded sequence in $X$, and $A$ is compact so there is some subsequence $(x_{n_k})$ such that $(A(x_{n_k}))$ converges in $Y$. So we have shown that any sequence in $A(S_X)$ has a convergent subsequence, but to conclude that the set $A(S_X)$ is sequentially compact we must show that the limit of this subsequence lies in $A(S_X)$, which I have no idea how to do.
The reverse direction I'm stuck on also. Given a bounded sequence $(x_n)$ in $X$, provided there are no zeroes in this sequence, we have that $(x_n/||x||)$ is a sequence in $S_X$, so then by the sequential compactness of $A(S_X)$ its image has a convergent subsequence $(A(x_{n_k}/||x||)$. But then I have no idea how to get from the convergent subsequence $(A(x_{n_k}/||x||)$ to some convergent subsequence of $(A(x_n))$.
I also have no clue where the fact that $Y$ is Banach is important!! Any tips with any of these steps would be great 🙂
Best Answer
As stated, this is not true. Consider for instance $Y=L^2[0,1]$ and $X\subset Y$ with $X=\mathbb C[x]$, the polynomials. Let $$ (Tf)(x)=\int_0^x f(t)\, dt. $$ This is the well-known Volterra Operator, which is compact. Note that the image of $T$ consists of the polynomials $p$ with $p(0)=0$. Here the "sequentially" in the "sequentially compact" means nothing, because we are dealing with metric spaces. So the claim is that the image of the unit ball is compact, which in particular would imply it is closed. But the set of polynomials with norm at most 1 is not closed.