# Importance of the uniform boundedness principle

I've heard that the uniform boundedness principle from functional analysis is a quite important result.

The theorem is the following:

Let $$X$$ be a Banach space and $$Y$$ a normed vector space. Let $$F$$ be a collection of continuous linear operators $$T:X\to Y$$ and suppose that $$\sup_{T\in F}\|T(x)\|< \infty$$ for all $$x\in X$$, then

$$\sup_{T\in F}\|T\|=\sup_{T\in F, \|x\|=1}\|T(x)\|<\infty.$$

Now, what is the importance of this result? I really can't grasp why this principle is so important as I've seem people say.

My question here is: why is this principle so important, and what are the main important consequences of it?

• The point is that to show that $\sup_T \|T\|$ is bounded that you need only show that $\sup_T \|T(x)\|$ is bounded for each $x$. The equality above has nothing to do with the theorem as such. – copper.hat Nov 24 '16 at 20:52
• I know that the equality $\sup_{T\in F} \|T\| = \sup_{T\in F, \|x\|=1} \|T(x)\|$ has nothing to do with it. In truth, as I know, by definition $\|T\| = \sup_{\|x\|=1}\|T(x)\|$, so that equality is just the definition of $\|T\|$. My doubt here is really why $\sup_{T\in F}\|T(x)\|$ bounded implying $\sup_{T\in F}\|T\|$ bounded is important, and what are the consequences of this fact. – Gold Nov 24 '16 at 21:08
• As an analogy, in Banach spaces weakly bounded and strongly bounded are equivalent, but it is usually easier to prove a set is weakly bounded. – copper.hat Nov 24 '16 at 21:16
• Related post on Quora: What is the significance of the uniform boundedness principle? – Martin Sleziak Aug 17 '17 at 14:20
• This post is also related: Important applications of the Uniform Boundedness Principle. – Martin Sleziak Aug 18 '17 at 14:08

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.

• Even if the proof of the claim at the end of your post is difficult and it's not possible to reproduce it here, maybe you would be ble to add at least some reference. – Martin Sleziak Sep 1 '17 at 15:42
• @MartinSleziak Nice to see your interest! Give me some time and I will either provide a reference or write it up compactly and add it to the answer. – Franka Waaldijk Sep 1 '17 at 15:58
• Well...that was harder than I thought. I could not find a reference...so (noblesse oblige) I had to sit down and get the details right. Hopefully right, that is, please check. – Franka Waaldijk Sep 2 '17 at 12:46
• I dont see how you can say "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. " That is, I don't think we can just write $\lim_{n\to \infty}\|T_n(x)-T_n(w)\|< M\cdot \|x-w\| = \|\lim_{n\to \infty}(T_n(x)-T_n(w))\|< M\cdot \|x-w\| = \|T(x)-T(w)\|< M\cdot \|x-w\|$, because we only have pointwise convergence? So how can we draw your conclusions? – ManUtdBloke Sep 24 '17 at 15:45
• If there is $M \in \mathbb{R}$ such that: for all $x,w\in X$ and all $T\in F$: $\|T(x)-T(w)\|< M\cdot \|x-w\|$ Then $\|T^*(x)-T^*(w)\|< M\cdot \|x-w\|$ does hold for the pointwise limit $T^*$ as well. – Franka Waaldijk Sep 25 '17 at 9:19

Uniform Boundedness Principle, sometimes called Banach-Steinhaus Theorem, is one of the three "cornerstone" theorems in functional analyis; the other two are the Hahn-Banach Theorem and the Open Mapping Theorem. (Note that UBP and OMT each use the Baire Category Theorem in their proofs.) As copper.hat mentioned above, it states that to show $\sup_T\|T\|<\infty$, one need only show that $\sup_T\|Tx\|<\infty$ for some arbitrary $x$.

So, what exactly makes it such a cornerstone result? Well, it just comes up all the time in a great variety of proofs of other powerful results. For example, it is used to show that if a Schauder basis is subsymmetric (resp. symmetric) then it is uniformly subsymmetric (resp. symmetric). It is also used to show that if a linear operator is compact then so is its adjoint. A UBP argument shows that any weakly convergent sequence is norm-bounded, the Gelfand spectral radius formula $r(T)=\lim\|T^n\|^{1/n}$, Etc.

It would be impractical to compile a complete list, but the above examples are some good ones that immediately come to mind.

• You write: "it states that to show $\sup_T\|T\|<\infty$, one need only show that $\sup_T\|Tx\|<\infty$ for some arbitrary $x$", but that is incorrect. The principle states that to show $\sup_T\|T\|<\infty$, one need only show that $\sup_T\|Tx\|<\infty$ for all $x$. – Franka Waaldijk Sep 1 '17 at 12:19
• @FrankWaaldijk ... and presumably "for every $x$" is what he meant by "some arbitrary $x$" ... admittedly this usage is not very clear. – GEdgar Sep 1 '17 at 12:43
• Ah yes I see what you mean! I read it completely the way in which it was likely not meant ... :-) – Franka Waaldijk Sep 1 '17 at 12:56