application of the theorem of the open application

Question

Let $ X, Y $ be Banach spaces.

Suppose that $ T: X \to Y $ is a compact operator. show that if $ \dim Y $ is infinite, then $ T $ is not surjective.

idea: Using the theorem of the open application

Answer 1

Let $B_X$ (resp. $B_Y$) be the open unit ball of $X$ (resp. $Y$), and $\overline{B_X}$ (resp. $\overline{B_Y}$) its closure, the closed unit ball.

Assume $T$ is surjective and bounded. By the open mapping theorem, $T$ is open whence, in particular, $T(B_X)$ is an open neighborhood of $0$ in $Y$. So $T(B_X)$ contains $rB_Y$ (the open ball of radius $r$ centered at $0$) for some $r>0$. Then $$ rB_Y\subseteq T(B_X)\; \Rightarrow \; r\,\overline{B_Y}=\overline{rB_Y}\subseteq \overline{T(B_X)}\subseteq T(\overline{B_X})\; \Rightarrow\; \overline{B_Y}\subseteq r^{-1}T(\overline{B_X}). $$ Now if $T$ is compact, it means that $T(\overline{B_X})$ is relatively compact (compact, actually, in this case, but that's not needed). It follows that $\overline{B_Y}$ is compact, as a closed subset of the (relatively) compact set $r^{-1}T(\overline{B_X})$.

Hence $Y$ is finite-dimensional by (the application of) Riesz's lemma: a normed vector space is finite-dimensional if and only its closed unit ball is norm compact.