Note: Here are two examples of partial converses which could be convenient.
Partial converse in section 27 of A Hilbert Space Problem Book by Paul R. Halmos:
Every bounded subset of a Hilbert Space is weakly bounded.
We can read in section $27$: Uniform boundedness
The celebrated principle of uniform boundedness (true for all Banach spaces) is the assertion that a pointwise bounded collection of bounded linear functionals is bounded. The assumption and the conclusion can be expressed in the terminology appropriate to a Hilbert space $\mathbf{H}$ as follows.
The assumption of pointwise boundedness for a subset $\mathbf{T}$ of $\mathbf{H}$ could also be called weak boundedness; it means that foreach $f$ in $\mathbf{H}$ there exists a positive constant $\alpha(f)$ such that $|(f,g)|\leq \alpha(f)$ for all $g$ in $\mathbf{T}$.
The desired conclusion means that there exists a positive constant $\beta$ such that $|(f,g)|\leq \beta\|f\|$ for all $f$ in $\mathbf{H}$ and all $g$ in $\mathbf{T}$; this conclusion is equivalent to $\|g\|\leq \beta$ for all $g$ in $\mathbf{T}$.
It is clear that every bounded subset of a Hilbert space is weakly bounded. The principle of uniform boundedness (for vectors in a Hilbert space) is the converse: every weakly bounded set is bounded.
And later on in section $51$: Uniform boundedness of linear transformations
... the generalization of the principle of uniform boundedness from linear functionals to linear transformations is somewhat subtler. The generalization can be formulated almost exactly the same way as the special case: a pointwise bounded collection of bounded linear transformations is uniformly bounded. The assumption of pointwise boundedness can be formulated in a weak manner and a strong one.
A set $\mathbf{Q}$ of linear transformations (from $\mathbf{H}$ into $\mathbf{K}$) is weakly bounded if for each $f$ in $\mathbf{H}$ and each $g$ in $\mathbf{K}$ there exists a positive constant $\alpha(f,g)$ such that $|(Af,g)|\leq \alpha(f,g)$ for all $A$ in $\mathbf{Q}$. The set $\mathbf{Q}$ is strongly bounded if for each $f$ in $\mathbf{H}$ there exists a positive constant $\beta(f)$ such that $\|Af\|\leq \beta(f)$ for all $A$ in $\mathbf{Q}$.
It is clear that every bounded set is strongly bounded and every strongly bounded set is weakly bounded. The principle of uniform boundedness for linear transformations is the best possible converse.
Partial converse in section $18.2$ of Mathematical Methods in Physics: Distributions, Hilbert Space Operators, and Variational Methods by Philipp Blanchard and Erwin Brüning
Every uniformly bounded family of continuous linear functionals is pointwise bounded
We can read in section $18.2$: The weak Topology:
$\mathbf{Definition\ 18.2.3}$ Let $X$ be a Banach space with norm $\|\cdot\|$ and $\{T_\alpha:\alpha\in A\}$ a family of continuous linear functionals on $X$ ($A$ an arbitrary index set). One says that this family is
- $\mathbf{pointwise\ bounded}$ if, and only if, for every $x\in X$ there is a real constant $C_x<\infty$ such that
$$\sup_{\alpha\in A}|T_\alpha(x)|\leq C_x$$
- $\mathbf{uniformly\ bounded}$ or $\mathbf{norm\ bounded}$ if, and only if
$$\sup_{\alpha \in A}\sup\{|T_\alpha(x)|:x\in X,\|x\|\leq 1\}=C< \infty.$$
Clearly, every uniformly bounded family of continuous linear functionals is pointwise bounded.
For a certain class of spaces the converse is also true and is called the principle of uniform boundedness or the uniform boundedness principle.
Maybe also interesting is the following generalisation to Hausdorff locally convex topological vector spaces stated in Appendix C: The Uniform Boundedness Principle
$\mathbf{Definition\ C.1.1}$ Suppose $(X,\mathcal{P})$ and $(Y,\mathcal{Q})$ are two Hausdorff locally convex topological vector spaces over the field $\mathbb{K}$. Denote the set of linear functions $T:X\rightarrow Y$ with $L(X,Y)$. A subset $\Lambda\subset L(X,Y)$ is called
a) $\mathbf{pointwise\ bounded}$ if, and only if, for every $x\in X$ the set $\{Tx:T\in\Lambda\}$ is bounded in $(Y,\mathcal{Q})$, i.e., for every semi-norm $q\in \mathcal{Q}$,
$$\sup\{q(Tx):T\in\Lambda\}=C_{x,q}<\infty;$$
b) $\mathbf{equi}$-$\mathbf{continuous}$ if, and only if, for every semi-norm $q\in\mathcal{Q}$ there is a semi-norm $p\in\mathcal{P}$ and a constant $C\geq 0$ such that
$$q(Tx)\leq Cp(x)\qquad\qquad\forall x\in X, \forall T\in\Lambda.$$
Obviously, the elements of an equi-continuous family of linear mappings are continuous and such a family is pointwise bounded. For an important class of spaces $(X,\mathcal{P})$ the converse holds too, i.e., a pointwise bounded family of continuous linear mappings $\Lambda\subset L(X,Y)$ is equi-continuous.
Note: The conclusion of uniform boundedness on bounded sets in the classical uniform boundedness principle for normed spaces is equivalent to the condition that the family of operators is equicontinuous with respect to the original topology of the domain spaces. See e.g. this paper by Ronglu Li and Charles Swartz.
To understand the importance of the result, it helps to clarify that the statement
($\ast$) For all $x\in X$ there is $M_x\in\mathbb{R}$ such that for all $T\in F$: $\|T(x)\|< M_x$
is apparently much much weaker than the statement
($\ast\ast$) There is an $M \in \mathbb{R}$ such that for all $x\in X, T\in F$: $\|T(x)\|< M$
since ($\ast\ast$) tells us that the $x$-dependent bound $M_x$ (for $\sup_{T\in F}\|T(x)\|$) does not depend on $x$ at all, and can be chosen uniformly.
The Uniform Boundedness Principle (UPB) tells us that ($\ast$) implies ($\ast\ast$)!
Since:
($\ast$) $\ \ \Leftrightarrow$ $\ \sup_{T\in F}\|T(x)\|< \infty$ for all $x\in X$
($\ast\ast$) $\,\,\Leftrightarrow$ $\ \sup_{T\in F}\|T\|=\sup_{T\in F, \|x\|=1}\|T(x)\|<\infty$
The importance of this can be compared to situations where continuity implies uniform continuity (like on compact spaces), or where pointwise convergence of functions implies overall convergence (in some metric).
In fact, we find an important result right from those examples:
If for a family $F$ of bounded linear operators $T: X\rightarrow Y$ we know:
$\sup_{T\in F}\|T(x)\|< \infty$ for all $x\in X$
then (stronger than that every $T$ in $F$ is uniformly continuous, which we knew) we can conclude that there
is $M \in \mathbb{R}$ such that:
for all $x,w\in X$ and all $T\in F$: $\|T(x)-T(w)\|< M\cdot \|x-w\|$.
This in turn implies, with $X, Y$ as above:
($\star$) If $(T_n)_{n\in\mathbb{N}}$ is a sequence of pointwise converging continuous linear operators from $X$ to $Y$, then the linear operator $T$ defined by $T(x):= \lim_{n\in\mathbb{N}} T_n(x)$ is again continuous.
Someone who is not impressed with the theorem that continuity implies uniform continuity for compact spaces, could easily think that ($\star$) is not the most striking consequence of UBP.
However, just to give an impression of how strong a tool topology is, let me say that ($\star$) is actually equivalent to UBP. (The proof of which is not difficult but timespace-consuming, so I omit it.).
A second reason why UBP is useful stems from the fact that it can be much harder to calculate operator norms than vector norms. Thus, UBP can help a great deal in situations where we want to show (but cannot calculate directly) that certain operators $T\in F$ are bounded, by showing/calculating $\sup_{T\in F}\|T(x)\|< \infty$ for all $x\in X$ instead.
Since a proof that ($\star$) is equivalent to UBP is asked for, I will add it here. We only need to show that ($\star$) implies UBP, meaning that from ($\star$) and ($\ast$) we must deduce ($\ast\ast$).
$\newcommand{\yclos}{\tilde{Y}}$ So let $X,Y$ as above. We will use ($\star$) for operators from $X$ to the norm-completion $\yclos$ of $Y$. So we know:
($\star$) If $(T_n)_{n\in\mathbb{N}}$ is a sequence of pointwise converging continuous linear operators from $X$ to $\yclos$, then the linear operator $T$ defined by $T(x):= \lim_{n\in\mathbb{N}} T_n(x)$ is again continuous.
and we assume that we have a family $F$ of continuous linear operators from $X$ to $Y$ satisfying:
($\ast$) For all $x\in X$ there is $M_x\in\mathbb{R}$ such that for all $T\in F$: $\|T(x)\|< M_x$.
We will prove:
($\ast\ast$) There is an $M \in \mathbb{R}$ such that for all $x\in X, T\in F$: $\|T(x)\|< M$
Proof:
Suppose NOT ($\ast\ast$). Then there are sequences $(U_n)_{n\in\mathbb{N}},(w_n)_{n\in\mathbb{N}}, (M_n)_{n\in\mathbb{N}}$ where for all $n\in\mathbb{N}$:
(i) $U_n\in F$, $w_n\in X$ and $M_n=\sup_{T\in F}\|T(w_n)\|\in\mathbb{R}$
(ii) $\|U_0\|>2^{2}$ and $\|U_n\|>2^{2}\cdot M_{n-1}\cdot \|U_{n-1}\|$ for $n>0$
(iii) $\|U_n\|-\|U_n(w_n)\|<1$
(I leave it as an exercise to show these sequences exist).
From (ii),(iii) we deduce
(iv) $M_n > 2^{2}\cdot M_{n-1}$.
We know that for all $x\in X$ there is $M_x\in\mathbb{R}$ such that for all $T\in F$: $\|T(x)\|< M_x$. So we know that the operator $S$ from $X$ to $\yclos$ defined by
$S(x):= \Sigma_n 2^{-n}\cdot \frac{1}{M_{n-1}}\cdot U_n(x)$
is actually a well-defined pointwise-limit linear operator.
By ($\star$) we know that $S$ is continuous, therefore bounded. Yet we claim, for all $n\in\mathbb{N}$:
claim: $\|S(w_n)\|>2^{n}$
proof:
a) For $i<n$ we know that $\frac{1}{M_{i-1}}\cdot \|U_i(w_n)\| < 2^{-4}\cdot\frac{1}{M_{n-1}} \cdot\|U_n(w_n)\|$. Therefore
$$\|\Sigma_{i<n} 2^{-i}\cdot \frac{1}{M_{i-1}}\cdot U_i(w_n)\| < 2^{-3}\cdot\frac{1}{M_{n-1}}\|U_n(w_n)\|$$
b) For $n<i$ we know that $\frac{1}{M_{i-1}}\cdot \|U_i(w_n)\| < 1$ since $M_{i-1}\geq M_n=\sup_{T\in F}\|T(w_n)\|$. Therefore:
$$\|\Sigma_{n<i} 2^{-i}\cdot \frac{1}{M_{i-1}}\cdot U_i(w_n)\| < 1$$
c) $\|U_n(w_n)\| > 2\cdot M_{n-1}\cdot \|U_{n-1}\|$ and $\|U_{n-1}\|> 2^{2n}$ so
$$\|2^{-n}\cdot \frac{1}{M_{n-1}}\cdot U_n(w_n)\| > 2\cdot 2^n$$
By a), b), and c):
$$\|S(w_n)\| > 2\cdot 2^n - 2^{n-2} - 1 > 2^n$$
which contradicts that $S$ is bounded. From this contradiction, we conclude ($\ast\ast$).
And finally, we conclude that ($\star$) implies, and therefore is equivalent to, UPB.
Best Answer
To illustrate the point which Davide explained in his answer, let us have a look at a concrete example.
Let us take $E=$ the set of all real sequences with finite support (i.e., only finitely many terms are non-zero). Let us use the norm $\|x\|=\sup_n |x_n|$. The space $E$ is a linear normed space, but it is not a Banach space. (So the assumptions of Banach-Steinhaus theorem are not fulfilled.)
Let us take $F=\mathbb R$ and $T_n(x)=\sum_{k=1}^n x_k$.
For every $x\in E$ we have $|T_n(x)| \le \sum_{k=1}^n |x_k|$, which is a finite number. So $\sup_n |T_n(x)|<+\infty$ for any fixed $x\in E$.
But if we take $x_n=(\underset{\text{$n$-times}}{\underbrace{1,\dots,1,}}0,0,\dots)$, then $T_n(x_n)=n$ and $\|x_n\|\le 1$. So we see that $T_n(x)$ is not bounded on the unit ball, i.e. $$\sup_{\|x\|\le 1, n\in\mathbb{N}} T_n(x)=+\infty.$$