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Let $ X $ be Hilbert space, and $X'$ be its dual. Assume, $T:X' \rightarrow X$ is linear, continuous and $N: X \rightarrow X'$ is nonlinear, continuous and uniformly bounded. Does it imply that $T \circ N : X \rightarrow X$ is compact?

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What does it mean for a nonlinear function to be compact? –  copper.hat Oct 3 '12 at 15:52
    
@copper.hat Same as for linear: bounded sets are mapped to sets with compact closure. This property appears in Schauder's fixed point theorem. –  40 votes Jul 14 '13 at 3:15
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No, unless $X$ is finite dimensional. Identify $X$ with its dual $X'$. Take $T$ to be the identity and $N(x)=x$ if $\|x\|\leq 1$ and $N(x)=x/\|x\|$ otherwise. Then, $N$ is "nonlinear", continuous and uniformly bounded and the composition $x\mapsto T(N(x))$ is not compact: $TN$ maps the unit ball in $X$ onto itself and this is not compact unless $X$ is finite dimensional.

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