4
$\begingroup$

In an arbitrary normed space $X$ consider a sequence of functions $f_n:X\to \mathbb{R}$ which converges uniformly to some function $f: X\to \mathbb{R}$. Now consider an arbitrary function space $Y$ (over $\mathbb{R}$) consisting of functions from $X$ to $\mathbb{R}$ with arbitrary norm $\|\cdot\|_Y$, which contains $f_n$ for each natural $n$ and contains $f$.

Now intuitively I feel that uniform convergence should imply that $\|f-f_n\|_Y \xrightarrow[\infty]{n}0$. It is easy enough to prove this in any concrete function space like the $L^p$ spaces, but I'm struggling to find a proof for the general case, which is frustrating because it seems intuitively so obvious.

I feel the proof should go along the lines of, because for any $\varepsilon>0$ we can find a natural $N$ such that for all $n>N$ $$|f(x)-f_n(x)|<\varepsilon \quad \forall x\in X,$$ then in some sense $f-f_n$ is in some sense within an $\varepsilon$ "distance" of $0$. I'm struggling to make this argument concrete using only the ideas of general normed spaces. I now believe that there exists a strange norm which makes this untrue. Translated to metric spaces obviously the discrete metric would give us a suitable counterexample. I can't find a suitable analogue in a normed space because of scalar multiplication.

If anyone can give a proof or provide a counterexample as to whether uniform convergence implies convergence in the norm, or can direct me to a reference on the topic I'd be very appreciative.

EDIT: I'd like to rephrase the question to deal exclusively with the kind of spaces I had in mind, which Daniel Fischer uncannily knew. As he pointed out a counter example will be any space $L^P(X)$ where $\mu(X)=\infty$. So let us rephrase the question and deal with a compact subspace of $X$, say $Z$. Then if we consider a sequence $f_n:Z\to \mathbb{R}$ converging to $f:Z\to \mathbb{R}$, and redefine $Y$ to be a function space consisting of functions from $Z$ to $\mathbb{R}$ with some arbitrary norm, does uniform convergence then imply convergence in the norm? As my measure theory course was rather disappointing I'm not entirely sure that the measure of a compact subset is finite. If not then obviously the answer stays the same. Are there any conditions that force the statement to be true?

$\endgroup$
  • 2
    $\begingroup$ Look again at your proof for $L^p$ spaces. You probably assumed finite measure of the whole space, didn't you? $\endgroup$ – Daniel Fischer Sep 30 '16 at 13:08
  • $\begingroup$ @DanielFischer Yes I did in fact. It's obvious now that my proof doesn't work with infinite measure of the whole space. Please see my edit for a rephrasing. Thanks for the pointer. $\endgroup$ – K.Power Sep 30 '16 at 14:00
  • $\begingroup$ For a special case of measures called Radon, you can say that the measure of a compact set is finite $\endgroup$ – KyleW Sep 30 '16 at 14:33
  • $\begingroup$ Are you saying that if we have uniform convergence, then we have convergence in every other $L^p$ norm? i.e. uniform convergence implies convergence in every norm? $\endgroup$ – Pinocchio Dec 10 '17 at 19:51
2
$\begingroup$

This is not going to work with any norm. For instance if $X=[0,1]$, $Y=C^1(X)$ with the norm $$||f||_Y=\sup_{x\in [0,1]}\big(|f(x)|+|f'(x)|\big)$$ then uniform convergence of $f_n$ to $f$ (where $f_n$ and $f$ are $C^1$) does not ensure that $||f_n-f||_Y\rightarrow 0$.

$\endgroup$
  • $\begingroup$ Could you please give an example where uniform convergence does not imply convergence in the norm? $\endgroup$ – K.Power Sep 30 '16 at 14:40
  • 1
    $\begingroup$ Take $f_n(x)=\dfrac{\sin(2n\pi x)}{\sqrt{n}}$. Then $f_n$ converges uniformly to $f(x)=0$ on $[0,1]$ because $|\sin(2n\pi x)|\leq 1$. But $f_n'(x)=2\pi\sqrt{n}\cos(nx)$ does not even converge pointwise to $f'(x)$. $\endgroup$ – Olivier Moschetta Sep 30 '16 at 15:02
  • $\begingroup$ Thanks so much. As this is quite a simple example then I think there aren't any general constraints we can impose on our vector space to ensure the result holds. It all depends on the norm it looks like. $\endgroup$ – K.Power Sep 30 '16 at 15:44

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.