Prove that $(X,\Vert {\cdot}\Vert_2)$ is not complete where $\Vert f\Vert_2=\left(\int_{-2}^{2}|f(t)|^2 dt\right)^{1/2}$

Question

Prove that $(X,\Vert {\cdot}\Vert_2)$ is not complete where $X=C[-2,2]$ and \begin{align}\Vert f\Vert_2=\left(\int_{-2}^{2}|f(t)|^2 dt\right)^{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}\;-2\leq t\leq 0,\\nt,& \text{if}\;0\leq t\leq \frac{1}{n},\\1 &\text{if}\;\frac{1}{n}\leq t\leq 2.\end{cases} \end{align} Clearly, the above function is continuous by Pasting Lemma. Next, we show that the function is Cauchy.

Let $m,n\in \Bbb{N}$ such that $m\geq n.$ Then, we show that \begin{align} \Vert f_n-f_m\Vert_2=\left(\int_{-2}^{2}|f_m(t)-f_n(t)|^2 dt\right)^{1/2} \to 0,\; \text{as}\;n\to \infty.\end{align}

Now, \begin{align} \Vert f_n-f_m\Vert_2\leq\left(\int_{-2}^{2}|f_m(t)||f_m(t)-f_n(t)| +|f_n(t)||f_m(t)-f_n(t)| dt\right)^{1/2} \end{align}

We know that \begin{align} \int_{-2}^{2}|f(t)| dt\leq\sqrt{4}\left(\int_{-2}^{2}|f(t)|^2 dt\right)^{1/2}\end{align}

So, \begin{align} \Vert f_n-f_m\Vert_2\leq\left(\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)| dt\right)^{1/2} \end{align}

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

Answer 1

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

Answer 2

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 $0$.