# Prove that L 2 PC[−1, 1] is not a complete normed space

I'm trying to prove that the normed space of all piecewise continuously functions with the norm $$\int^1_{-1}|f(x)|^2dx$$ is not a complete normed space.

$L_2PC[-1,1]$

for that, im trying to find a Cauchy sequence function which does not converge in this space with this norm. (maybe the Cauchy sequence converges, but to a function which doesn't belong to $L_2PC[-1,1]$)

I have been stuck on this for a while, and I coule use some help. Thanks

• You could try approximating the indicator function $\mathbf 1_{[0,1/2]}$ by piecewise continuous functions (piecewise linear should work). Show that it is a Cauchy sequence (shouldn't be too hard). Then you will have a Cauchy sequence that does not converge in your space. – a... Jan 17 '15 at 22:35
• I fail to see how this function series helps me, can you please be a little more specific? – user3921 Jan 17 '15 at 23:22

## 1 Answer

Consider the sequence $(f_n)_{n \geq 0}$ with $$f_n(x) = \sum_{k=0}^n 2^{-k} \chi_{[1-2^{-k+1},\, 1-2^{-k})}(x)$$ where $\chi$ denotes the indicator function. If I got the indices right these should be step functions where $f_{n+1}$ differs to $f_n$ by a new step of length $2^{-(n+1)}$ and height $2^{-(n+1)}$. Every $f_n$ is piecewise continuous and they converge in $L^2$ to $$f(x) = \sum_{k=0}^\infty 2^{-k} \chi_{[1-2^{-k+1},\, 1-2^{-k})}(x),$$ but this limit has an infinite number of discontinuities.

• I can see how every $fn$ is piecewise continuous and I can also see how this sequence is a Cauchy sequence, I "see" the steps functions...but I do not see a a limit function with infinite number of discontinuities. I see this sequence converges to $f(x)=0$ poinwise and that's it....can you please clarify? thanks alot – user3921 Jan 18 '15 at 7:40
• The steps are disjoint and the step height doesn't depend on $n$, so for $I_k = [1-2^{-k+1}, 1-2^{-k})$ there holds $f_n|_{I_k} = 2^{-k}$ for all $n \geq k$. The same is true for the limit, so $f(x) \neq 0$ for $x \neq 1$. The step heights tend to zero for $k \to 0$, but this only means that $f(x) \to 0$ for $x \to 1$. – Three.OneFour Jan 18 '15 at 8:16