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I've been trying to come up with an easy example of an application of the uniform boundedness principle (or Banach-Steinhaus theorem). I was thinking of something like the following, which is unfortunately a non-example:


Non-example:

Consider the Banach space $(B(X), \|\cdot\|_\infty)$, i.e. the space of bounded functions with the sup norm. Let's choose $X=\mathbb R$ and let $\mu$ be a measure that is finite on compact sets, like for example the Lebesgue measure. Define $$ T_t : (B(X), \|\cdot\|_\infty) \to \mathbb R$$ as $$ f \mapsto \int_{[-t,t]} f d \mu$$

Then $T_t$ is linear and bounded: If $\|f\|_\infty = 1$ then $\|T_t f\| = |T_t f| = \int_{[-t,t]} f d \mu \leq 2t$ so that the operator norm $\|T_t\| \leq 2t < \infty$. The condition that fails that prevents me from applying Banach-Steinhaus is that the family $\{T_t\}_{t \in \mathbb R}$ is not pointwise bounded: For $f$ with $\|f\|_\infty$ we have $\sup_{t \in \mathbb R} \|T_t\| \geq \sup_{t \in \mathbb R} 2 t = \infty$.


My non-example was supposed to conclude that the integral operator is continuous on the whole space. Of course it's not on $\mathbb R$ for bounded functions but you see what I'm trying to do. Although one might consider showing that the integral is a continuous operator by applying Banach-Steinhaus "cracking nuts with a sledge hammer", this one is for educational purposes so it's acceptable.

Can someone either modify my example so that it works or show me an equally easy example? Thanks a lot for your help.

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    $\begingroup$ You should remove the absolute value around f in the definition of $T_t$ to get it linear. $\endgroup$
    – agt
    Commented Jul 29, 2012 at 10:24
  • $\begingroup$ Oh, of course, otherwise I don't have $T(f+g) = Tf + Tg$ but only $\leq$! Thank you @GiuseppeTortorella $\endgroup$ Commented Jul 29, 2012 at 10:26
  • $\begingroup$ And what if you replace $\chi_{[−t,+t]}\mu$ with $\frac{1}{t\sqrt{2\pi}}e^{−x^2/2t^2}\mu$ defining $T_t$? $\endgroup$
    – agt
    Commented Jul 29, 2012 at 11:20
  • $\begingroup$ Dear @GiuseppeTortorella, so you mean $$f \mapsto \int_{\mathbb R} f(x) \cdot \frac{1}{t\sqrt{2\pi}}e^{−x^2/2t^2} d x$$? $\endgroup$ Commented Jul 29, 2012 at 11:34
  • $\begingroup$ This reminds me of a Fourier or Laplace transform, does this have a name? $\endgroup$ Commented Jul 29, 2012 at 11:35

2 Answers 2

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Here's an easy example of Banach-Steinhaus (BS)(perhaps calling it Uniform Boundedness Principle makes it slightly easier to memorise since it contains more information):

We can use BS to show that $\mathbb R [x]$, the space of polynomials over $\mathbb R$, is not a Banach space with respect to the following norm: $\|p(x)\| = \|\sum_{k=0}^N a_k x^k \| := \max_k |a_k|$.

Define a bounded linear operator $T_n : \mathbb R [x] \to \mathbb R$, $p(x) \mapsto \sum_{k=0}^n a_k$. It's linear (easy to see) and it's bounded since the sum is finite.

Also, for a fixed point $p(x) \in \mathbb R [x]$, $$ \|T_n p \| = \left |\sum_{k=0}^n a_k \right | \leq (\mathrm{deg}p + 1) \|p\|$$ that is, the family $\{T_n\}$ is pointwise bounded. Hence if $\mathbb R [x]$ was a Banach space, all the requirements for BS would be fulfilled so that by BS, $\sup_{n \in \mathbb N} \| T_n \|$ should be finite. But this is false since we can define $p_n (x) = 1 + x + \dots + x^n$ so that $\|T_np_n \| = n + 1 $ and hence $\|T_n\| \geq n + 1$ and hence $\sup_{n \in \mathbb N} \|T_n \| = \infty$. This would contradict BS, hence $\mathbb R [x]$ must be incomplete.


Edit (in response to comment where t.b. kindly pointed out that my $p_n$ is not Cauchy):

This seems like cracking nuts with a sledgehammer since one can just observe that the sequence $p_n(x) = \sum_{k=0}^n \frac{x^k}{2^k} $ has an infinite sum as its limit which is by definition not a polynomial.

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    $\begingroup$ How does $p_n$ have an infinite sum as its limit with respect to your norm? It's not even Cauchy! $\endgroup$
    – t.b.
    Commented Jul 29, 2012 at 16:05
  • $\begingroup$ @t.b. Thanks a lot for spotting the mistake. Ok, got me, I ballsed up there. But then how about $$p_n(x) = \sum_{k=0}^n \frac{1}{2^k} x^k$$ That is Cauchy and its limit is also not in the space. : ) $\endgroup$ Commented Jul 29, 2012 at 16:13
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    $\begingroup$ yeah, that works :) Also your estimate $\lVert T_n p\rVert \leq \lVert p \rVert$ is slightly off: you should multiply the right hand side by $\deg{p}+1$. $\endgroup$
    – t.b.
    Commented Jul 29, 2012 at 16:15
  • $\begingroup$ @t.b. Oops, true dat. Thanks for spotting that, too! : ) $\endgroup$ Commented Jul 29, 2012 at 16:16
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    $\begingroup$ Remember this :) $\endgroup$
    – t.b.
    Commented Jul 29, 2012 at 18:13
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Here's another slightly more important example:

We can use Banach-Steinhaus to show that even though we have convergence in $L^2$ for Fourier series of $f \in C(\mathbb T)$, convergence does not hold pointwise. This we can show by showing that there exists an $f \in C(\mathbb T)$ such that the Fourier series of $f$ diverges at $0$.

For this we need the following three facts (proofs omitted):

Fact 1: If $D_n = \sum_{k=-n}^n e^{2 \pi i k}$ is the $n$-th Dirichlet kernel then $$ D_n \ast f(x) = \sum_{k=-n}^n \hat{f}_k e^{ik 2 \pi x}$$ that is, the convolution with the Dirichlet kernel gives us the $n$-th partial sum of the Fourier series of $f$.

Fact 2: $$ \int_0^1 \left | D_n(x) \right | dx \xrightarrow{n \to \infty} \infty$$

Fact 3: The map $T_n: f \mapsto D_n \ast f (0)$ is a bounded linear operator with norm $\|T_n\| = \int_0^1 \left | D_n(x) \right | dx$.

Now all the prerequisites for Banach-Steinhaus are fulfilled:

(i) $(C(\mathbb T), \|\cdot\|_{L^2})$ is a Banach space, $\mathbb R$ is a normed space

(ii) $\{T_n\}$ is a family of bounded linear operators: for each $n$, $T_n$ is bounded by $\|T_n\|$ (Fact 3)

Now if $\{T_n\}$ was pointwise bounded, i.e., for a given $f$, $\sup_{n \in \mathbb N} \|T_n f \| < \infty $ then we could apply Banach-Steinhaus to get $\sup_{n \in \mathbb N} \| T_n \| < \infty$ but that would be a contradiction to $\| T_n \| = \int_0^1 |D_n(x) | dx$ and $ \int_0^1 \left | D_n(x) \right | dx \xrightarrow{n \to \infty} \infty$. Hence $\{T_n\}$ cannot be pointwise bounded that is, there exist an $f$ such that $T_nf$ diverges for $n \to \infty$.

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