Does closedness of the image of unit sphere imply the closed range of the operator

Question

Let $X$ and $Y$ be Banach spaces and let $T:X\to Y$ be a bounded linear operator such that $T(S_X)$ is closed in $Y$. Does it imply that $T(X)$ is closed? Any hint is appreciated.

Answer 1

The answer is no in general (but it's not difficult to check that the answer is yes for injective operators).

Counterexample. Let $H$ be a Hilbert space and $T_0: H \to H$ a bounded linear operator with non-closed range. Consider the operator $T: H \times H \to H \times H$ given by $T(x,y) = (T_0 x, 0)$ for all $x,y \in H$, where $H \times H$ is endowed with, say, the usual norm induced by $H$ that turns $H \times H$ into a Hilbert space. The range of $T$ is $(T_0 H) \times \{0\}$, so it is not closed.

However, let us now show that $T(S_{H \times H}) = T_0(B_H) \times \{0\}$, where $B_H$ denotes the closed unit ball in $H$. The inclusion "$\subseteq$" is clear. For the converse inclusion "$\supseteq$", take $x \in B_H$. Then one can choose $y \in B_H$ such that $(x,y) \in S_{H \times H}$ and hence $(T_0 x, 0) = T(x,y) \in T(S_{H \times H})$.

Since $H$ is reflexive, the set $B_H$ is weakly compact. As $T_0$ is continuous with respect to the weak topology, it follows that $T_0(B_H)$ is also weakly compact, hence weakly closed, and thus closed. Therefore, $T(S_{H \times H}) = T_0(B_H) \times \{0\}$ is closed in $H \times H$.

Remark. Of course, a similar construction can also be done with other reflexive Banach spaces.

Answer 2

Functional analysis is awash with pairs of Banach spaces where the first one is continuously embedded into the second one in such a way that the unit ball of the first one is closed in the second one, so much so that they are the basis of their own theory—that of mixed topologies or two normed space, prominent enough to have its own MOS classification section (46A70). Probably their first use was in Saks‘ seminal paper on what was to be known as the Vitali-Hahn-Saks theorem, where he relies on the fact that the unit ball of $L^\infty$ is complete in the $L^1$-norm—hence the use of the name ��Saks spaces“.