Interchange closure and continous mapping in Banachspaces and alternative proof for Open mapping theorem

Question

Let $X,Y$ be Banach spaces and $T:X \to Y$ a continous and linear Operator. Let $U_1$ be the unit open ball in $X$. Is it true that $\overline{T(U_1)}=T(\overline{U_1})$?

"$\supseteq$" is clear.

"$\subseteq$" my reasoning is as follows:

It holds that $T(U_1)$ is dense in $\overline{T(U_1)}$. Thus $\forall y \in \overline{T(U_1)}\exists (y_n)_{n \in \mathbb{N}} \subset T(U_1): y_n \to y, \ n \to \infty$ Now for all $y_n$ it exists $x_n$ s.t. $y_n=Tx_n$ with $||x_n||<1$. Since $T$ is continous we have $y = T\lim\limits_{n \to \infty} x_n$. Since $X$ is complete we have an $x\in X$ with $Tx=y$. Since the norm is continous we have $||x||\leq 1 $ and therefore $y \in T(\overline{U_1})$

Is this reasoning correct? If not skip the following and please share the made mistake

I came across this question trying to understand a proof for the open mapping theorem. This one consisted of two steps:

1. Step: Show that $\exists \varepsilon >0: V_{\varepsilon} \subseteq \overline{T(U_1)}$ where $V_{\varepsilon}$ is the open $\varepsilon-$ball in $Y$
2. Step: Show that $\exists \varepsilon_0 >0: V_{\varepsilon_0} \subseteq T(U_1)$

Now I could follow the first step but the second step seemed to me too complicated and too long. Doesn't step 2 follows trivially from step 1 and the lemma above from the fact that:

$y \in V_{\varepsilon} \Rightarrow y=Tx$ for some $x$ with $||x||\leq 1$ (lemma above). Thus $\alpha y=T\alpha x$ with $||\alpha x||<1$ for $\alpha \in (0,1) \Rightarrow V_{\alpha \varepsilon} \subseteq T(U_1)$

Answer 1

Your statement is not true. Here is an example (below is also an explanation of a mistake in your proof). Consider $X = C[-1,1]$ and $Y = L_1(-1,1)$. There is a canonical embedding $T:C[0,1] \rightarrow L_1(0,1)$. I claim that $T(\overline{U}_1)$ is not equal to $\overline{T(U_1)}$. Consider $f_n(x) = \frac{2}{\pi}\arctan(nx) \in C[-1,1]$. It is obvious that $f_n \in U_1$ i.e. $||f_n|| < 1$. But in space $L_1(-1,1)$ sequence $f_n$ converges to $f(x) = sgn(x)$, i.e. $$ f(x) = \begin{cases} 1 & \mbox{if }x > 0 \\ -1& \mbox{if }x < 0 \end{cases} $$ $f$ is not continuous and therefore $f \notin T(C[-1,1])$ and in particular $f \notin T(\overline{U_1})$.

In your proof you made a mistake when you claimed that sequence $x_n$ (a sequence s.t. $Tx_n = y_n$) is converging. You said that since $X$ is complete it converges. It appears that $x_n$ can be not a Cauchy sequence. In example from above sequence $f_n$ is a Cauchy sequence in $L_1(-1,1)$ but it is not a Cauchy sequence in $C[-1,1]$.