Universal properties of mapping spaces in functional analysis Recently I've been trying to learn more about functional analysis, and have been wondering about what universal characterisations are there of mapping spaces for topological vector spaces, (semi)normed spaces, and Banach spaces.
In particular, I've seen the following characterisation invoked a few times:

The operator norm $||{-}||$ on $\mathcal{Ban}(X,Y)$ with $X$ normed and $Y$ Banach is the unique one such that

*

*$\mathcal{Ban}(X,Y)$ is also a Banach space.

*For any sequence $\{T_{n}\}_{n\in\mathbf{N}}$ of operators, if $\displaystyle\lim_{n\to\infty}(||T_{n}||)=0$, then $\displaystyle\lim_{n\to\infty}(T_{n}(x))=0$ for all $x\in X$.

So if $||{-}||'$ is another norm on $\mathcal{Ban}(X,Y)$ satisfying those conditions, then it must be the case that $||{-}||$ and $||{-}||'$ are equivalent.


*

*What would be a reference (or proof) for the above?

*Are there other similarly nice such "universal characterisations" of $\mathcal{Ban}(X,Y)$?

*Lastly, are there analogous results as the above one for semi/normed spaces and topological vector spaces?

 A: To answer your first question, this characterization of the operator norm is a consequence of the uniform boundedness principle and the open mapping theorem.  Specifically, suppose you have a norm $\|\cdot\|'$ such that for each $x\in X$, the operator $ev_x:\mathcal{Ban}(X,Y)\to Y$ defined by evaluation at $x$ is bounded (this is equivalent to your condition (2)).  Note that if you restrict $x$ to lie in the unit ball of $X$, then these operators $ev_x$ are pointwise bounded.  So if $\|\cdot\|'$ is complete, the uniform boundedness principle implies the operators $ev_x$ for $\|x\|\leq 1$ are uniformly bounded.  This means there exists $C\in\mathbb{R}$ such that for all $x\in X$ and $f\in\mathcal{Ban}(X,Y)$ with $\|x\|\leq 1$ and $\|f\|'\leq 1$, $\|ev_x(f)\|=\|f(x)\|\leq C$.  That is, if $\|f\|'\leq 1$, then the operator norm $\|f\|$ is at most $C$.
So, we have shown that if $\|\cdot\|'$ is another norm satisfying (1) and (2), then the identity map $(\mathcal{Ban}(X,Y),\|\cdot\|')\to (\mathcal{Ban}(X,Y),\|\cdot\|)$ is bounded.  Since both norms are complete, the open mapping theorem implies the inverse map is also bounded, so the two norms are equivalent.
(The uniform boundedness principle and open mapping theorem do have generalizations to certain topological vector spaces beyond Banach spaces, so I would expect there is some generalization of this characterization to a broader class of topological vector spaces.  I don't know the details off the top of my head, though.)
