Htykuut

Prove that $(X,Vert {cdot}Vert_2)$ is not complete where $X=C[-2,2]$ and
begin{align}Vert fVert_2=left(int_{-2}^{2}|f(t)|^2 dtright)^{1/2}.end{align}

MY TRIAL

It suffices to produce a Cauchy sequence in $C[-2,2]$ that does not converge to a point in $C[-2,2]$.

Consider the function defined by

begin{align} f_n(t)=begin{cases}0,& text{if};-2leq tleq 0,\nt,& text{if};0leq tleq frac{1}{n},\1 &text{if};frac{1}{n}leq tleq 2.end{cases} end{align}
Clearly, the above function is continuous by Pasting Lemma. Next, we show that the function is Cauchy.

Let $m,nin Bbb{N}$ such that $mgeq n.$ Then, we show that
begin{align} Vert f_n-f_mVert_2=left(int_{-2}^{2}|f_m(t)-f_n(t)|^2 dtright)^{1/2} to 0,; text{as};nto infty.end{align}

Now, begin{align} Vert f_n-f_mVert_2leqleft(int_{-2}^{2}|f_m(t)||f_m(t)-f_n(t)| +|f_n(t)||f_m(t)-f_n(t)| dtright)^{1/2} end{align}

We know that begin{align} int_{-2}^{2}|f(t)| dtleqsqrt{4}left(int_{-2}^{2}|f(t)|^2 dtright)^{1/2}end{align}

So,
begin{align} Vert f_n-f_mVert_2leqleft(int_{-2}^{2}|f_m(t)||f_m(t)-f_n(t)|dt +int_{-2}^{2}|f_n(t)||f_m(t)-f_n(t)| dtright)^{1/2} end{align}

I'm stuck here and don't know how to proceed. Any help please?

You need to use that $f_n(t)-f_m(t)=0$ if $tle 0$ or $tge frac1n$. In $[0,frac1n]$, you have $|f_n(t)-f_m(t)|le 1$. So
$$
int_{-2}^2 |f_n(t)-f_m(t)|^2 , dt le frac1nlongrightarrow 0.
$$

Hint: Use $(int_{-2}^{2} |f_n(t)-f_m(t)|^{2})^{1/2} leq (int_{0}^{2} |f_n(t)-1|^{2})^{1/2}+(int_{0}^{2} |f_m(t)-1)|^{2})^{1/2}$. It is a lot easier to prove that each of the two terms tend to $0$.