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I was reading Rudin's Principles of Mathematical Analysis, and I came across the definition 7.19, where it says that a sequence of functions $f_n(x)$ is pointwise bounded on E if there exists a finite-valued function $\phi$ defined on E such that $$ |f_n(x)| < \phi(x) $$ for x element of E, n = 1, 2 ,3 ... While $f_n$ is uniformly bounded on E if there exists a number M s.t. $|f_n(x)| < M$ for x element of E, n = 1, 2 , 3 ... But if we define the set U as the values of $\phi(x)$ from our first definition and define the sup of the set as R, then don't we get the second definition. Wouldn't that mean that pointwise bounded implies uniform bounded?

Rishi
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    The issue is that the sup of U may be infinite. Remember, no one said that $\phi$ has to be bounded. – Jesse Madnick Nov 17 '10 at 01:32
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    I don't follow. If $f_n(x) < \phi(x)$ then every element in U is finite, since $\phi(x)$ is a finite valued function, no? – Rishi Nov 17 '10 at 01:33
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    Yes, every element in U may be finite, and $\phi(x)$ is finite-valued, but $\phi$ can still be unbounded. – Jesse Madnick Nov 17 '10 at 01:34
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    For example, let $E = (0,1)$ and let $\phi(x) = 1/x$. In fact, let each $f_n(x) = 1/x$. This sequence is pointwise bounded by $\phi$ on $(0,1)$, and certainly finite at each point of $(0,1)$, but $\phi$ is unbounded. – Jesse Madnick Nov 17 '10 at 01:35
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    Pointwise boundedness means that for EACH $x_0 \in E$, the sequence ${f_n(x_0)}$ is a bounded sequence of real numbers. So, if all of the $f_n$'s are the same thing (for example), then for each $x_0$, the sequence ${f_n(x_0)}$ will be a constant sequence, hence bounded. But again, if we let the $x_0$'s vary, then what this constant sequence is might be arbitrarily large. – Jesse Madnick Nov 17 '10 at 01:38
  • Another problem with Rudin's definition is that "finite-valued" was never defined. – Dick Grayson Jan 23 '22 at 14:51

2 Answers2

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The short answer is: there need not be a real number that is the supremum of the values of $\phi(x)$. You may have $\sup\{\phi(x)\mid x \in E\} = \infty$. If that is the case, you're out of luck.

Arturo Magidin
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This an excellent example of a subtle issue that keeps coming up. The property of being pointwise bounded ensures that for every element of your domain of definition the sequence of functions is bounded at that point. Where as uniform boundedness says that there exists an upper bound that holds for every element of your domain.

I decided to answer this old question because I had a very similar thought regarding The Principle of Uniform Boundedness in my Functional Analysis class.
Another point worth mentioning is that this is yet another case where the quantifiers do not commute. Another classical example being Uniform Convergence vs Pointwise Convergence

Nick Castillo
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