# A sequence of functions $\{f_n(x)\}_{n=1}^{\infty} \subseteq C[0,1]$ that is pointwise bounded but not uniformly bounded.

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:

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!

You had the right idea, but don't let the spike have a gentle slope on the right. Try $$f_n(x) = \cases{n^2 x & if x < 1/n\cr n^2 (2/n - x) & if 1/n \le x \le 2/n\cr 0 & otherwise\cr}$$

Consider the $C^\infty[0,1]$ functions $f_n$ defined by $$f_n(x)=n^2x^n(1-x).$$ Then $f_n$ is maximum at $x_n=n/(n+1)$ and $f_n(x_n)\sim n/\mathrm e$ hence $\|f_n\|_\infty\to\infty$ when $n\to\infty$, but, for every $x$ in $[0,1]$, $f_n(x)\to0$ when $n\to\infty$ hence the sequence $(f_n(x))_{n\geqslant0}$ is bounded.

Look at these functions: $$f_n(x)=\begin{cases}2 n^2 x \ & \textrm{ if } 0 \leq x \leq \frac{1}{2n} \\ n-2n^2x & \textrm{ if } \frac{1}{2n} < x \leq \frac{1}{n} \\ 0 & \textrm{ otherwise}\\ \end{cases}.$$

Note that for every $0<x<1$ we have $f_n(x)=0$ if $\frac{1}{n}<x$, so, there are only finite n's such that $f_n(x) \neq 0$ and this family is pointwise bounded.

In the other hand, it's not uniformly bounded, because the maximum of the function $f_n$ is $n$.

Hope I have not committed any error, sorry if I did something wrong.

First, note that your pointwise bound $\varphi(x)$ must be discontinuous, else it would be bounded on $[0,1]$ and you'd have a uniform bound. Now, consider the function $$f(x)=\begin{cases}\frac 1 x&x\neq 0\\{}\\ 0&x=0\end{cases}$$

We know this is unbounded near $x=0$. At each $1,1/2,\ldots$ this admits a tangent line that is strictly below the function. This tangent is $$T_n(x)=f(n^{-1})+f'(n^{-1})\left(x-\frac 1n\right)\\=2n-n^2x$$

This certainly will follow $f$s behaviour. Then, we continuously join this tangent point to the origin, using $$T'_n(x)=n^2x$$

and to keep $T_n$ positive we cut it off at its root, $2n^{-1}$. Note this gives Robert's solution, and indeed this idea will work for any convex function that goes unbounded near the endpoints.