# How to prove that convergence is equivalent to pointwise convergence in $C[0,1]$ with the integral norm?

I'm trying to prove (or disprove) that in the set $C[0,1]$ of continuous (bounded) functions on the real interval [0,1] with the integral norm $\|f(x)\|_1 = \int_0^1|f(x)|dx$ that a sequence of functions $f_n$ is convergent to $f$ if and only if $f_n$ is pointwise-convergent to $f$.

My intuition tells me that this is true, but I'm having trouble formulating an attack (particularly for the convergence => pointwise convergence case). Given $\forall \epsilon > 0 \; \exists N \in \mathbb{N} \;$ $\forall n \geq N \; \int_0^1 |f_n(x) - f(x)|dx < \epsilon$, I'm trying to prove that for an arbitrary $x \in [0,1]$, $\forall \epsilon_0 > 0$ there is an $N_0$ such that for $n_0 \geq N_0 |f_{n_0}(x) - f(x)| < \epsilon_0$. I think I want to make $N_0$ somehow depend on the that integral (or a 'slice' of it), but I don't see it yet. Any pointers/advice/hints are welcome.

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Sometimes if a proof is not working out, it pays to start looking for counterexamples. –  Byron Schmuland Feb 8 '11 at 2:29
Dear Bosmacs, this is false (as @Byron hints above). You may consider a sequence of step functions of successively larger height but rapidly decreasing width, whose integrals tend to zero but which do not pointwise converge to zero. You may then approximate them by continuous functions. –  Akhil Mathew Feb 8 '11 at 2:34
Also, observe that you can simplify the problem: the statement is true if and only if it works when $f$ is the constant zero function. –  Jim Belk Feb 8 '11 at 2:35
@Jim I don't quite see why that's the case. Can you elaborate? –  bosmacs Feb 8 '11 at 2:47
I think Jim Belk was referring to the fact that both notions of convergence are invariant under translation. Specifically, $f_n\to f$ (either in $1$-norm or pointwise) if and only if $f_n-f\to 0$ (in the same sense). –  Jonas Meyer Feb 8 '11 at 3:44

## 3 Answers

It is possible for a sequence of continuous functions to converge in the $1$-norm to a continuous function, while converging pointwise nowhere.

It is also possible for a sequence of continuous functions to converge pointwise everywhere to a continuous function without converging in $1$-norm.

The last claim is the easier of the 2 to verify. Consider functions whose graphs are spikes of rapidly increasing height with base on say $[0,1/n]$.

For the first claim, you can consider a more complicated sequence of spikes, again getting narrower, but this time not increasing in height, and this time have them move all over the place (but in an orderly enough fashion to ensure the lack of convergence at any point).

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Consider $\displaystyle f_n(x) = \begin{cases} nx & \text{ for } 0 \leq x \leq \tfrac{1}{2n} \\ 1-nx & \text{ for } \tfrac{1}{2n} \lt x \lt \tfrac{1}{n} \\ 0 & \text{ otherwise}\end{cases}.$

Clearly $\lVert f_n\rVert_1$ is constant (the integral is just the area of a triangle), but $f_n \to 0$ pointwise and $(f_n)$ isn't convergent w.r.t. $\lVert \cdot \rVert_\infty$.

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Check the beginning of the second paragraph. –  Shai Covo Feb 8 '11 at 3:52

Hint: Consider $f_n(x) = 2xn^2$ if $0 \leq x \leq 1/(2n)$, $f_n(x) = 2n - 2xn^2$ if $1/(2n) \leq x \leq 1/n$, and ...

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