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?

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    $\begingroup$ 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

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 $||f_n-f||_Y\rightarrow 0$.

  • $\begingroup$ 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
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    $\begingroup$ 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

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