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?
Real Analysis – Why Pointwise Bounded Does Not Imply Uniform Bounded
analysisreal-analysis
Related Solutions
Comparison
Pointwise convergence means at every point the sequence of functions has its own speed of convergence (that can be very fast at some points and very very very very slow at others).
Imagine how slow that sequence tends to zero at more and more outer points:
$$\frac{1}{n}x^2\to 0$$
Uniform convergence means there is an overall speed of convergence.
In the above example no matter which speed you consider there will be always a point far outside at which your sequence has slower speed of convergence, that is it doesn't converge uniformly.
Another Approach
One can check uniform convergence by considering the "infimum of speeds over all points". If it doesn't vanish then it is uniformly convergent. And that gives another characterization as the ones with nonvanishing overall speed of convergence.
A bounded function is a function whose range is contained in a finite interval. A sequence of bounded functions has the property that each of its individual functions $f_n$ has such a limited range (but the overall range of the collection may still be infinitely large).
The sequence is pointwise bounded if for every $x$ separately the sequence $f_n(x)$ is contained in a finite interval. That is a different condition.
Best Answer
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.