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?
Why doesn't pointwise bounded imply uniform bounded?
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4The issue is that the sup of U may be infinite. Remember, no one said that $\phi$ has to be bounded. – 2010-11-17
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0I 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? – 2010-11-17
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0Yes, every element in U may be finite, and $\phi(x)$ is finite-valued, but $\phi$ can still be unbounded. – 2010-11-17
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6For 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. – 2010-11-17
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4Pointwise 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. – 2010-11-17
1 Answers
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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.