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There's like three applications of the uniform boundedness principle in wikipedia:

1) If a sequence of bounded operators converges pointwise to an operator, then the limit operator is also bounded, and the convergence is uniform on compact sets.

2) Any weakly bounded subset of a normed space is bounded.

3) A result in pointwise convergence of Fourier series.

I am just asking if there's more interesting applications of the uniform boundedness principle.

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    $\begingroup$ It can be used to show that at most finitely many of coefficient functionals corresponding to a Hamel basis can be continuous; see this post. $\endgroup$ – Martin Sleziak Jul 6 '14 at 8:04
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Let $f\in L^{p}(\mathbb{T})$, for some $1\leq p<\infty$, where $\mathbb{T}$ denotes the one-dimensional torus. Let $(a_{m})_{m\in\mathbb{Z}}$ be a bounded complex sequence. For $R\geq 0$, let $(a_{m}(R))_{m=1}^{\infty}$ be a compactly supported sequence such that $a_{m}(R)=a_{m}$ for all $\left|m\right|\leq R$. Define $$S_{R}(f)(x):=\sum_{m\in\mathbb{Z}}a_{m}(R)\widehat{f}(m)e^{2\pi im\cdot x},\ \forall x\in\mathbb{T}$$ For each $R$, the above expressions defines an operator $L^{p}\rightarrow L^{p}$ that maps a function $f$ to the $\lfloor{R}\rfloor^{th}$ symmetric partial sum of its Fourier series.

Using, in part, a simple application of the Uniform Boundedness Principle, we can turn the question of $L^{p}$ convergence of $S_{R}(f)$ to $f$ as $R\rightarrow\infty$, for arbitrary $f\in L^{p}$, into a question of the uniform boundedness of the operators $S_{R}$, $R\geq 0$.

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Here's a couple more examples.

If $f : \Omega\subseteq\mathbb{C} \rightarrow \mathcal{L}(X)$ is a function from an open subset $\Omega$ of the complex plane into the bounded linear operators $\mathcal{L}(X)$ on a complex Banach space $X$, then $f$ is holomorphic iff $\lambda\mapsto x^{\star}(f(\lambda)x)$ is holomorphic on $\Omega$ for all $x \in X$, $x^{\star}\in X^{\star}$.

If $f$ is a holomorphic vector function on the punctured disk $0 < |\lambda| < \delta$ with values in a Banach space $X$, then $f$ has an essential singularity at $0$ iff there exists $x^{\star}\in X^{\star}$ such that $x^{\star}\circ f$ has an essential singularity at $0$.

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  • $\begingroup$ Thanks, can I have a reference where I can find the proofs ? $\endgroup$ – user50618 Jul 6 '14 at 22:58
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    $\begingroup$ Many texts on Functional Analysis deal with the first issue, including Rudin. You can start by lifting from $\lambda \mapsto x^{\star}(f(\lambda)x)$ being holomorphic to $\lambda \mapsto f(\lambda)x$ being holomorphic. Then you can lift again to $\lambda\mapsto f(\lambda)$ being holomorphic. As for the second part, I think that's actually an application of the Baire category theorem, which is used to establish uniform boundedness. $\endgroup$ – Disintegrating By Parts Jul 8 '14 at 0:03
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One other application I know is in the proof of the spectral radius formula: given a Banach space $X$ and $T\in B(X)$ a bounded linear opeartor, $$ \operatorname{spr}(T)=\lim_{n\to\infty}\|T^n\|^{1/n}, $$ where $\operatorname{spr}(T)=\max\{|\lambda|:\ \lambda\in\sigma(T)\}$.

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