We were talking about pointwise bounded vs. uniformly bounded in my analysis class, and this question came up. The problem is that we are working on a compact set, it would be much easier if the interval was $(0, 1]$. My idea was to create a sequence of functions such that $f_n(\frac{1}{n}) = n$ and $f_n(0) = 0$, $f_n(1) = 0$ and then connect the "spike" with line segments to the endpoints. Visually, the $f_n's$ would look like mountains. After working out the slopes, I came up with these formulae:
$$f_n(x) = \left\{
\begin{array}{ll}
n^2x & \quad 0 \leq x \leq \frac{1}{n} \\
\frac{-n^2}{n-1}(x-1) & \quad \frac{1}{n} < x \leq 1
\end{array}
\right.$$
This gives the picture I was visualizing in my head (unless my arithmetic is incorrect), but unfortunately, it does not work, since it is not pointwise bounded. My idea was that at $x = \frac{1}{m}$ $f_n(x) \leq m$ for all $n$. But this is not the case, for example, $f_{10}(\frac{1}{2}) = \frac{200}{9} \geq 5$.
So the question is: can you give me a sequence of functions $\{f_n(x)\}_{n=1}^{\infty} \subseteq C[0,1]$ that is pointwise bounded but not uniformly bounded? And if so, is there anyway to save my construction? There must be a "canonical" example, because otherwise the uniform boundedness in the conclusion of the Arzela-Ascoli Theorem would not really be relevant.
I did search for answers to this question, and did not find any. I found these:
Equicontinuity implies (pointwise bounded iff uniformly bounded)
Why doesn't pointwise bounded imply uniform bounded?
Does this problem make sense? "Give an example of a set $F\subset C([0,1])$ which is pointwise bounded but not bounded"
The last one is obviously the same question I am asking, but it has no answer, and I could not think of anything based off the hint.
Thanks!
Best Answer
You had the right idea, but don't let the spike have a gentle slope on the right. Try $$ f_n(x) = \cases{n^2 x & if $x < 1/n$\cr n^2 (2/n - x) & if $1/n \le x \le 2/n$\cr 0 & otherwise\cr}$$