This questions relates to nonlinear compact operators. That is an operator $D$ on a Hilbert space $H$ with $D: H\rightarrow H$ with $D$ defined to be compact on $S\subset H$ if $D$ is continuous and if for every bounded set $V \subset S$, $D(S)$ is a relatively compact set. (For the purposes of this question I'm assuming $H$ is real and only considering the case of an operator on a bounded set V rather than the more general case).
Now suppose we also have a nonlinear scalar function $f:H \rightarrow \mathbb{R}$ and that $f$ is continuous and that f(V) is a bounded set in $\mathbb{R}$ with $V$ a bounded set in $H$.
Is the operator f(x)Dx (that is the scalar function multipled by the operator) also a compact operator on $V$?
I've come up with a 'proof' (given below). But I can't help feeling I'm missing something here. Any help on whether my proof is actually correct or whether this result is simply a corrollary of another result or indeed false would be very gratefully received.
Proof Let $Tx:=f(x)Dx$. Then the continuity of $T$ follows immediately. Let $v_n$ be an arbitary sequence in $V$. Since $D$ is compact there exists a subsequence $v_{n_i}$ for which $Dv_{n_i}$ converges. Now consider the sequence $f(v_{n_i})$ by Bolzano-Weierstrass this has a convergent subsequence $f(v_{n_{i_j}})$. That is we are now considering a sub-subsequence. The sub-subsequence $Dv_{n_{i_j}}$ must also converge. Hence our operator $Tv_{n_{i_j}} =f(v_{n_{i_j}}) Dv_{n_{i_j}} $ must converge. So T(V) is relatively compact and thus T is a compact operator.