convergence with respect to integral norm but not pointwise

I want to give an example of a sequence of functions $f_1 \dots f_n$ that converges with respect to the metric $d(f,g) = \int_a^b |f(x) - g(x)| dx$ but does not converge pointwise.

I'm thinking of a function $f_n$ that is piecewise triangle, whose area converges to some constant function, but doesn't converge pointwise.

I just can't manage to formalize it.

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Think of the most basic functions you know first. It helps. See my answer. =) –  Patrick Da Silva Apr 15 '12 at 20:36
There is a standard example for this, think about the unit interval with a square function that slides back and fourth with the squares area going to 0 by means of it's width shrinking (keep height same). –  Alex R. Apr 15 '12 at 20:38
Bleh, I misread your question. Sorry for the silly comment/answer. I've been computing differential equations for too long for one of my exams, and I began entering in robot mode / stop thinking.. –  Patrick Da Silva Apr 15 '12 at 21:39
I think the whole idea is just to move the area around fast enough so that pointwise convergence is not obtained. –  Patrick Da Silva Apr 15 '12 at 21:40
I edited my answer. Take a look, it should be fine now =) –  Patrick Da Silva Apr 15 '12 at 21:43

Since you can always replace $f$ by $f-g$, you might as well start by assuming that $g(x)=0$. Then you can use your idea by making part of the graph of $y=f_n(x)$ a triangle of height $1$ and base $2^{-n}$, say, the rest lying on the $x$-axis. For instance, you might have $$f_1(x)=\begin{cases}0,&\text{if }0\le x\le \frac12\\2x-1,&\text{if }\frac12\le x\le 1\;.\end{cases}$$ If you make the triangles peak alternately at $1$ and $0$, you’ll ensure that the $f_n$ don’t converge pointwise.

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Here's an example of a sequence, mentioned by Sam in the comments, of a sequence that converges pointwise nowhere on $[0,1]$ but $\int_{[0,1]} f_n\rightarrow 0$:

Let

$\ \ f_1=\chi_{[0,1]}$, $f_2=\chi_{[0,{1\over2}]}$, $f_3=\chi_{[{1\over2},1]}$, $f_4=\chi_{[0,{1\over4}]}$, $f_5=\chi_{[{1\over4},{2\over4}]}$, $f_6=\chi_{[{2\over4},{3\over4}]}$, $f_7=\chi_{[{3\over4},1]}$, $\ldots$.

Here $\chi_A$ is the function whose value is 1 on the set $A$ and $0$ on the set $A^C$.

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The easiest answer would be to think of a sequence of functions $f_n : [0,1] \to \mathbb{R}$

$$f_n(x)=\left\{\begin{matrix} 1 \ \ \ \ \ \ \ x\neq 0 \\ (-1)^n \ x=0 \end{matrix}\right.$$

which converges to the constant function $1$ w.r.t the integral norm. But it does not converge point-wise to a function as $(-1)^n$ does not converge.

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your sequence converges a.e to the constant function 1. –  toufik_kh.17 Apr 16 '14 at 5:45

You can suppose that $g(x) = 0$, because your metric is translation invariant ; $d(f_n,g) = d(f_n-g,0)$. Think of the sequence $f_n : [0,1] \to \mathbb R$ defined by $f_n(x) = x^n$ if $n$ is odd, and $(1-x)^n$ if $n$ is even. Therefore, $$d(f_{2n+1},0) = \int_0^1 (1-x)^{2n+1} \, dx = \frac 1{2n+2} \underset{ n \to \infty} {\longrightarrow} 0$$ or $$d(f_{2n},0) = \int_0^1 x^{2n} \, dx = \frac 1{2n+1} \underset{ n \to \infty} {\longrightarrow} 0$$ but $f_n$ does not converge pointwise at $0$ and $1$ because the values oscillates between $0$ and $1$.

I know my answer uses the idea of peaking "the triangles" alternatively at $0$ and $1$ of Brian, but I still think it's worth it to see a more-or-less "trivial" example (using not-defined-by-parts polynomial functions), so I kept my answer there anyway.

Hope that helps,

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I think that the OP is looking for a sequence which does not converge pointwise. –  Xabier Domínguez Apr 15 '12 at 20:45
@Xabier : Thanks for noticing. I edited my answer. –  Patrick Da Silva Apr 15 '12 at 21:44