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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!

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5 Answers 5

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

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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.

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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.

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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.

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Pick any sequence $X=\{x_n\}$ from $[0,1]$. For the sequence of functions $F=\{f_n\}$ to be pointwise bounded on $X$ would mean the sequence $$f_1(x_n)\:\:\:f_2(x_n)\:\:\: f_3(x_n) \:\:. . .$$ to be bounded for every $n$. For each of the functions to be bounded on $X$ would mean the sequence $$f_n(x_1)\:\:\: f_n(x_2)\:\:\: f_n(x_3)\:\:. . .$$ to be bounded for every $n$. Keeping this in mind I construct the following double sequence : \begin{matrix} 1&1&1&1&\dots\\ 1&2&2&2&\dots\\ 1&2&3&3&\dots\\ 1&2&3&4&\dots\\ \vdots&\vdots&\vdots&\vdots&\vdots \end{matrix}

Since each row and each column is bounded (eventually constant) we have a pointwise bounded sequence of bounded functions. But since the diagonal is $(1,2,3,4,...)$ the sequence is not uniformly bounded.

Continuity can be achieved very easily as well. For instance :

on (0,1]

Of course this only works on $(0,1]$ since the sequence is not pointwise bounded on $0$. To include $0$ some modification is required :

enter image description here

Here the $nth$ function reaches the top at $\frac{1}{n}$ and starts declining at $\frac{1}{2^n}$. The flat top of the functions will forever keep pushing to the left, never reaching zero; exhibiting the pointwise boundedness of the sequence.

Notice how the pointwise limit on $(0,1]$ is the same in both the cases.

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