image of convex closed, bounded subspace of H by monotonous continuous operator is closed Let H be a Hilbert space and K a convex closed, bounded subspace of H. Let $F : H → H$ be continuous monotonous function. Prove that $F(K)$ is closed. My argument is the following: K is closed under H so K is a Hilbert space. F is continuous and monotonous so F(K) is a Hilbert subspace of H. I am not sure how to go from there. Any hints or help? Thank you.

This is not true for $K$ being a subspace. Take $F$ to be linear, compact, and monotone. Let $K=H$. Compact operators do not have closed range.

It is true, if $K$ is assumed to be a convex, closed, bounded subset of $H$. Let $(x_n)$ be a sequence in $K$ such that $F(x_n)\to y$. It remains to show that there exists $x\in K$ with $F(x)=y$.

Since $K$ is weakly sequentially compact, we can assume (after possibly extracting a subsequence) that $x_n \rightharpoonup x$ with $x\in K$. Take some $v\in H$. By monotonicity, we have $$ \langle F(x_n)-F(v),x_n-v\rangle\ge0. $$ Passing to the limit shows $$ \langle y-F(v),x-v\rangle\ge0\quad \forall v. $$ Setting $v=x+ tw$ for some $w\in H$, $t>0$ gives $$ \langle y-F(x+ tw),-t w\rangle\ge0 $$ Dividing by $t$, letting $t\searrow 0$ gives $$ \langle y-F(x),-w\rangle\ge0\quad \forall w $$ hence $y=F(x)$.