$T:X \to Y$ bounded linear map and $X$ separable implies $Y$ is separable?

Let $T:X \to Y$ be a bounded linear map between Banach spaces. Suppose that $X$ is separable.

Is it true that $Y$ has to be separable?

I think yes, since the map is continuous it takes the countable dense subset of $X$ to a countable dense subset of $Y$.

• I presume you want $T$ to be onto? – David Mitra Jul 7 '15 at 12:00
• Yes, let it be bijective. – QuestionAnswer Jul 7 '15 at 12:13
• Surjective is enough. Your last statement will then be true. – David Mitra Jul 7 '15 at 12:15

This is true in a more general setting. Let $f: X\to Y$ be a continuous map between two topological spaces and $Q \subset X$ a dense subset. Then $f(Q)$ is dense in $f(X)$.
To see this, let $y\in f(X)$. Then let $x\in X$ so that $f(x) = y$. Let $V \subset Y$ be open and $y\in V$. Then $U = f^{-1}(V)$ is open and contains $x$. As $Q$ is dense in $X$, there is $q\in Q$ so that $q\in U$. Thus $f(q) \in V\cap f(X)$. As $y, V$ are arbitrary, this shows that $f(Q)$ is dense in $f(X)$.
• Yes, it's true. The answer will be a bit long to be written here, but the main observation is that if $T: X \to Y$ is a bounded bijective linear map between two Banach space, then so are $T^*$ and $T^{**}$. That $T^{**}$ is bijective would force the canoncial embedding $Y \to Y^{**}$ to be surjective. – user99914 Jul 7 '15 at 13:20
• Concerning reflexivity you can also argue with the weak* compactness of the closed unit balls. If $T:X\to Y$ is onto then the unit ball of $Y$ is contained in a multiple of $T(B_X)$ (by the open mapping theorem) which is weakly compact. Hence $B_Y$ is relatively weakly compact and weakly closed, hence weakly compact. – Jochen Jul 7 '15 at 13:25