# Supremum of a strong-weak continuous map under compact set

Let $T:X \to Y$ be a map between Hilbert spaces. Let $K$ be a compact set in $X$. If $T$ is continuous we know that $\sup_{k \in K} \lVert T(k)\rVert_Y$ is achieved for some element of $K$.

Suppose $T$ is not continuous but satisfies: $x_n \to x$ in $X$ implies $Tx_n \rightharpoonup Tx$ in $Y$ (weak convergence). Could the same result still hold regarding the compactness? Or anything like it?

• If $k \in K$, then $T(k) \in Y$. Then $\sup_{k \in K} T(k)$ is without any sense ! – Fred Mar 8 '17 at 10:35
• @Fred Oops. Let me fix – Upin Mar 8 '17 at 10:37

Suppose $T$ is not continuous but satisfies: $x_n \to x$ in $X$ implies $Tx_n \rightharpoonup Tx$ in $Y$ (weak convergence).
This means that $T$ is norm-weak continuous (as opposed to norm-norm continuous), i.e. continuous w.r.t to the norm in $X$ and the weak topology on $Y$.
There is the general result for continuous maps, that the image of a compact set is compact again, thus $T(K) = \{Tx \colon x \in K\}$ is weakly compact in $Y$, since $T$ is not norm-norm but norm-weak continuous.
The set $T(K)$ will be bounded and closed (see here), but it appears that in the mentioned case there will be no element in $T(K)$ achieving the supremum. See here for a counterexample.