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?
Best Answer
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 $||f_n-f||_Y\rightarrow 0$.