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It's well-known that when $X$ is a normed space and $Y$ is Banach space, the set $L(X,Y)$ of bounded linear maps between the spaces is a banach space with the norm $\|A\| = \sup_{\|x\|_{X}=1}\|Ax\|_{Y}$.

The proof of this can also be used to show a slightly more general result. $X$ can be a seminormed space instead of a normed space and the end result is still Banach:

Theorem: Let $(X,\rho)$ be a seminormed space and $Y$ a Banach space. Define $\|A\|_{\rho} = \sup_{\rho(x) \leq 1}\|Ax\|_{Y}$. Let $L_{\rho}(X,Y)$ be the space of linear maps that are bounded w.r.t $\rho$ (that is, $\|A\|_{\rho} < \infty$), with the norm $\|\cdot\|_{\rho}$. Then $L_{\rho}$ is a Banach space.

I am looking for a reference of this fact to include in a paper. Thanks.

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This is quite straight-forward. If $\varrho$ is a semi-norm, and $$ M=\{x\in X:\varrho(x)=0\}, $$ then $M$ is a linear subspace, and $X/M$ is naturally equipped with the norm $$ \|x/M\|=\varrho(x). $$ Also, if $x\in M$, then $Ax=0$. Otherwise, $A$ would not be bounded, in the sense described in the formulation of the Theorem.

Then, from $A:X\to Y$ we define, also naturally, the bounded linear transformation $$ \hat A: X/M\to Y, $$ where $\hat A(x+M)=Ax$.

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  • $\begingroup$ So If I'm understanding you correctly, it's simply that the space I'm referring to is isomorphic to the space of bounded linear maps from $X/M$ to $Y$, already known to be Banach? $\endgroup$ – ttb Jun 21 '17 at 16:42
  • $\begingroup$ I do not know whether this specific Theorem was known to Banach. $\endgroup$ – Yiorgos S. Smyrlis Jun 21 '17 at 16:48
  • $\begingroup$ What I mean is that the space $L(X/M, Y)$ is known to be a Banach space. $\endgroup$ – ttb Jun 21 '17 at 16:53
  • $\begingroup$ It is indeed well-known that $L(X/M,Y)$ is a Banach space, since $X/M$ is a normed space. $\endgroup$ – Yiorgos S. Smyrlis Jun 21 '17 at 16:55

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