$H$ is infinite-dimensional Hilbert space. $K: H \rightarrow H$ is infinite-dimensional compact operator. Let $S$ be a unit sphere in H. The task is to proof that $K(S)$ couldn't be compact.
I know that:
Really we need to proof that $K[S]$ is not closed. If B is unit ball, $[B]$ – closure, then $K([B])$ is also closed(for compact operator in Hilbert space).
If $\mathrm{Ker} B$ is empty, i.e. $0 \notin K(S)$ then there is a sequence in $K(S)$ converging to $0$, so $K(S)$ isn't compact. But I don't understant how to proof noncompactness of $K(S)$ even if one of the basis vectors is mapped to $0$.
On the other side we can try to proof that if $K(S)$ is compact, then $\mathrm{Im}(K)=K(H)$ shuold have finite dimension. It is easy if $K(B)$ contains some ball(like a subspace in $K(H)$), but it's far from being true.
Best Answer
In fact, $K(S)$ can be compact. Let $H = \ell^2(\mathbb{N})$, i.e. $$H = \Biggl\{ f \colon \mathbb{N} \to \mathbb{C} \biggm\vert \sum_{k = 0}^{\infty} \lvert f(k)\rvert^2 < +\infty\Biggr\}$$ (using function notation to make indexing of sequences in $H$ clearer) and $K = C \circ D$, where $$(Df)(k) = f(k+1)$$ and $$(Cf)(k) = 2^{-k}f(k)\,.$$ Then $K(S)$ is closed, hence compact.
Let $g \in \overline{K(S)}$, and $(f_n)_{n \in \mathbb{N}}$ a sequence in $S$ with $K(f_n) \to g$. If $g = 0$, then we have $K(e_0) = g$ where $e_m(k) = \delta_{m,k}$, hence we can assume $g \neq 0$. For every fixed $k$ we have $$f_n(k+1) \to 2^kg(k)$$ and therefore $$\sum_{k = 0}^{m} 2^{2k}\lvert g(k)\rvert^2 = \lim_{n \to \infty} \sum_{k = 0}^{m} \lvert f_n(k+1)\rvert^2 \leqslant 1$$ for all $m$. It follows that $h \in H$, where $$h(k) = 2^kg(k)\,.$$ Furthermore, we have $\lVert h\rVert \leqslant 1$, hence there is $h_1 \in S$ with $h = Dh_1$. By construction, $Ch = g$, thus $$g = Kh_1 \in K(S)\,.$$