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.
 A: 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 $A^C$.
A: 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.
A: 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.    
A: 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,
