0
$\begingroup$

I'm looking for a sequence of functions $\{f_n\}_{n\in\mathbb{N}}$, continuous on the interval $[a,b]$ such that it converges to $f$ under the $L^2$ norm, defined as $$\lVert \rVert_2=\left(\int_a^b (f_n(x)-f(x))^2dx\right)^{1/2},$$ but not under the $L^\infty$ norm, defined as $$\lVert \rVert_\infty=\sup_{x \in[a,b]}\lvert f_n(x)-f(x)\rvert.$$

I've tried a bunch of functions, especially trigonometric functions. However, every time, I get stumped by the power of two in the definition of the $L^2$ norm.

$\endgroup$
  • $\begingroup$ In this case, the limit is not in $L^2$; change it to $1/x^{1/4}$ and this works perfectly. EDIT: this was in response to a comment suggesting $f_n(x) = \max(n, 1/x)$ as a candidate (assuming the interval is $[0,1]$). $\endgroup$ – User8128 Mar 23 '18 at 17:27
1
$\begingroup$

Let $f_n(x) = x^n$ on $[0,1].$ Verify that $\|f_n - 0\|_2 \to 0,$ while $\|f_n - 0\|_\infty = 1$ for every $n.$

$\endgroup$
  • $\begingroup$ Wow, that's really simple, I can't believe it escaped me... Just to verify, in the $L^\infty$ norm, how can you claim the limit is $f=0$? This might seem like an elementary question, but is the limit unique regardless of the space's norm? $\endgroup$ – Grizzly0111 Mar 23 '18 at 18:43
  • $\begingroup$ @Grizzly0111 I do not understand your comment. I answered your question fully and precisely. $\endgroup$ – zhw. Mar 23 '18 at 18:57
  • $\begingroup$ Yes, you answered my question fully. You write $\lVert f_n - 0 \rVert _\infty = 1$. I just want to understand how you can claim that if the limit of $f_n$ exists in the space with the norm $\lVert \rVert _\infty$, then it is 0. My apologies if I was unclear. $\endgroup$ – Grizzly0111 Mar 23 '18 at 19:07
  • 1
    $\begingroup$ @Grizzly0111 Here's the question: "I'm looking for a sequence of functions ${f_n}$ continuous on the interval $[a,b]$ such that it converges to $f$ under the $L^2$ norm, but not under the $L^\infty$ norm." So my sequence $f_n(x)=x^n$ converges to $0$ in the $L^2$ norm, but does not converge to $0$ in the $L^\infty$ norm. Why aren't we done? $\endgroup$ – zhw. Mar 23 '18 at 19:41
  • $\begingroup$ Ah, my bad, you are right, we are done. $\endgroup$ – Grizzly0111 Mar 23 '18 at 19:43
3
$\begingroup$

Consider the constant function $f(x) = 1$ and let $f_n(x) = \chi_{[a+1/n, b]}$ where $\chi$ denotes the indicator functions. Then $$\|f - f_n\|_\infty = 1$$ for all $n$ since the functions differ by $1$ on the positive measure set $[a,a+1/n)$. However, it is easy to show that $f_n \to f$ in $L^p$ for any $p$.

As a side note, what you are really exploiting here is the fact that $L^p$ is separable for any $1\le p < \infty$ but $L^\infty$ is not separable.

EDIT: I notice now that we need $f_n$ to be continuous; this is no problem. Just change $f_n$ on the interval $[a, a+1/n)$ so that it is linear and satisfies $f_n(a) = 0, f_n(1/n) = 1$. The rest of the answer remains the same except now we may note that $$\| f_n -f\|_\infty \ge 1/2$$ since they differ by more than $1/2$ on a set of positive measure (of course, we still do have $\| f_n - f\|_\infty =1$ but you don't need to prove this).

$\endgroup$
  • 1
    $\begingroup$ the sequence $(f_n)$ needs to be continuous according to OP. Your sequence is not, it has jump discontinuities. $\endgroup$ – Arian Mar 23 '18 at 17:21
  • $\begingroup$ Thanks for the answer, it makes a lot of sense! One quick question to make sure I understand: when you say "Just change fn on the interval [a,a+1/n) so that it is linear and satisfies fn(a)=0,fn(1/n)=1 ", do you mean that I should take something like $f_n(x)=nx$ for 0<x<a+1/n, and $f_n(x)=1$ for a+1/n<x<b ? $\endgroup$ – Grizzly0111 Mar 23 '18 at 18:04
  • 1
    $\begingroup$ Yes, that’s exactly what I mean. But it should be $f_n(x) = n(x-a)$ I suppose. $\endgroup$ – User8128 Mar 23 '18 at 18:07
0
$\begingroup$

Try that $f_{1}=\chi_{[0,1]}$, $f_{2}=\chi_{[0,1/2]}$, $f_{3}=\chi_{[1/2,1]}$, $f_{4}=\chi_{[0,1/4]}$,...

$f_{n}\rightarrow 0$ in any $L^{p}[0,1]$ for $1\leq p<\infty$ but there is no any $x$ such that $\lim_{n\rightarrow\infty}f_{n}(x)$ exists.

To make them to be continuous, just like what @User8128 has done, join splines to the functions such that the areas of the triangles becoming more and more smaller.

$\endgroup$
0
$\begingroup$

Have $f_n(x)$ be zero on most of the interval except for a sharp quick spike that reaches $y=n$ on a subinterval short enough so that the integral is of order $O(\frac{1}{n})$.

Obviously $f_n \longrightarrow 0$ in $L^2$ norm, but $\|f_n-f_m\|_\infty $ is not small even for large $n$ and $m$, so $f_n$ cannot converge in this norm.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.