I run in to problem that I often know is solvable with either the Closed Graph Theorem or Uniform Boundedness Theorem. I seem to mix up the solutions. Are there any hints on when to use which? Or can they both be used solve the same problem?

One example: Let $X$ be a Banach space and let $T_n$ be a sequence of bounded linear maps from $X$ into itself, such that for every $x\in X$ we have $$ \lim_{n\rightarrow \infty} T_nx = x$$ in the norm of $X$. Show that the linear map $T:X\rightarrow X$ is continuous iff the maps $T_nT$ are continuous for each $n\geq 1$.

For ($\Leftarrow$), I tried to use the Closed Graph Theorem as follows. Assume $x_n \rightarrow x$ and $Tx\rightarrow y$. Then $$\lim_{m \rightarrow \infty} T_mT\lim_{n \rightarrow \infty}x_n = \lim_{m \rightarrow \infty} T_my = y,$$ and $$\lim_{m \rightarrow \infty} T_mT\lim_{n \rightarrow \infty}x_n = \lim_{m \rightarrow \infty} T_mTx = Tx,$$ and by the Closed Graph Theorem, we are done.

However, the solution to the exercise solved it using the Uniformed Boundedness Theorem.


1 Answer 1


The "big three" theorems about Banach spaces that occur frequently in functional analysis are:

  1. the Hahn-Banach Theorem (HBT),
  2. the Principle of Uniform Boundedness (PUB) (also known as the Uniform Boundedness Theorem or the Banach-Steinhaus Theorem), and
  3. the Open Mapping Theorem (OMT).

You could easily add two more "named theorems":

3(a). the Closed Graph Theorem (CGT), and

3(b). the Bounded Inverse Theorem (BIT).

However, $$\text{OMT} \iff \text{CGT} \iff \text{BIT}, \tag{1}$$ so as long as you remember that, you can reduce your mental list to the "big three" above.

PUB and OMT---although not equivalent---are siblings since they both come from the Baire Category Theorem. For more, see this.

Since you are specifically asking about CGT vs. PUB, it is worth stating (a version of) these side-by-side to compare and contrast:

Principle of Uniform Boundedness. Let $X$ be a Banach space and $Y$ a normed linear space. Let $\{T_n\}$ be a sequence of bounded linear operators, $T_n:X\to Y$ such that $\{\|T_nx\|\}$ is (pointwise) bounded, i.e., $$\exists C_x\text{(independent of $n$) such that } \|T_nx\|\le C_x, \qquad \forall x\in X,\ n\in\mathbb{N}.$$ Then $\{\|T_n\|\}$ is uniformly bounded, i.e., $$\exists C\text{ such that }\|T_n\|\le C, \qquad n\in\mathbb{N}.$$

Closed Graph Theorem. If $X$ and $Y$ are Banach spaces and $T:X\to Y$ is a closed linear operator (i.e., it has a closed graph), then $T$ is bounded.

Back to your question: if you are wondering if any problem you encounter that can be handled with one of these can also be handled with the other, the answer must be no since CGT requires the completeness of $Y$, whereas PUB does not.

However, assuming all the hypotheses of each theorem are satisfied, one can certainly envision problems (aimed at deducing the boundedness or continuity of an operator) where either could be used.

Keep in mind that HB doesn't require completeness of the underlying spaces (and hence typically appears first in texts); PUB requires $X$ (but not $Y$) to be complete; and OMT/CGT/BIT require both $X$ and $Y$ to be complete.

Think of the "big three" as related, but distinct tools for accomplishing (perhaps) different tasks. If you are dealing with linear functionals or dual spaces, HB should be on your radar. If you have pointwise estimates on a sequence of operators and want uniform estimates, look to PUB. If you want to deduce boundedness/continuity about an operator, look to the trio of OMT/CGT/BIT depending on what you know about the ingredients in that particular situation.

  • 2
    $\begingroup$ Great! thanks, it was a good read! $\endgroup$
    – Johan
    Jan 9, 2013 at 19:03
  • $\begingroup$ Excellent answer, ..deleted the rest of my comment as I see you answered it in your post. $\endgroup$
    – csss
    Mar 15, 2017 at 9:07
  • $\begingroup$ In fact, the PUB is equivalent to OMT/CGT/BIT. See here: mathoverflow.net/questions/190587/… $\endgroup$ Oct 4, 2021 at 21:30

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