7
$\begingroup$

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?

  • 4
    The issue is that the sup of U may be infinite. Remember, no one said that $\phi$ has to be bounded.2010-11-17
  • 0
    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?2010-11-17
  • 0
    Yes, every element in U may be finite, and $\phi(x)$ is finite-valued, but $\phi$ can still be unbounded.2010-11-17
  • 6
    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.2010-11-17
  • 4
    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.2010-11-17

1 Answers 1

8

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.