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this is an exercise from Brezis Book, 2.11. My attempt: T is surjective so thanks to the open mapping theorem, for every $n\in\mathbb{N}$ there exists $x_n\in E$ such that $\|x_n\|\leq c$ ($c$ is the constant from the open mapping theorem) and $T(x_n)=e_n$, where $(e_n)_{n\in\mathbb{N}}$ is the canonical basis of $\mathcal{l}^1$, and define the next function: $S:\mathcal{l}^1\rightarrow E$ where if $y=(y_n)_{n\in\mathbb{N}}\in\mathcal{l}^1$ then $$S(y)= \displaystyle{\sum_{n=1}^\infty y_n(x_n)}.$$ This "function" is near to be the right inverse of $T$ because $T\circ S=Id$, $S(y)$ is a convergent serie for every $y\in\mathcal{l}^1$ and $S$ is linear and bounded but I can't prove that is well defined, I mean, we don't know if $T$ is injective and maybe could exists an $n\in\mathbb{N}$ such that $e_n=T(x_1)=T(x_2)$ with $x_1\neq x_2$ so the definition of $S$ is ambiguous. Please could you explain me what is happening?

ifthenelse
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    The right inverse need not be unique, which is what you explain. But by "choosing" those vectors $x_n$ your formula produces a right inverse. Perhaps only one among many right inverses. You choose $x_n$ so that $T(x_n) = e_n$. Since $T$ is a function, and $e_1 \ne e_2$ we get $x_1 \ne x_2$. This is "$T$ is a function", and does not require "$T$ is injective". – GEdgar Mar 02 '21 at 00:34

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Let $T_n(y)=\sum_{i=1}^{i=n}y_nx_n$, $T_n$ is continuous and $\|T_n(y)\|\leq \|y\|c$.

We have $lim_nT_n(x)$ exists since $T_n(x)$ is a Cauchy sequence, the uniform bounded principle implies that there existsa bounded operator $T$ such that $T(x)=lim_nT_n(x)$. See the corrolaries

https://en.wikipedia.org/wiki/Uniform_boundedness_principle

Tsemo Aristide
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