Sequence Cauchy w.r.t $L^2$-norm but has no convergent subsequence I have a problem with following exercise: Let $C([0,1])$ be the space of continuous real valued functions on the closed unit inverval $[0,1]$. And let the $L^2$-norm given by: \begin{equation}
\Vert f\Vert_{L^2}:=\left(\int_0^1\vert f(t) \vert^2\right)^{1/2}
\end{equation}
Then I want to find a sequence of continuous functions $f_n:[0,1]\to \mathbb{R}$ that is Cauchy w.r.t. the $L^2$-norm and has NO convergent subsequence.
I though about following sequence: $f_n(x)=\frac{1}{n}\sin(2nx)$ for $x\in [0,\frac{\pi}{2}]$ and $f_n(x)=0$ otherwise. But first: I'm not sure if this works?! and Second, how can I check that there is no convergent subsequence? 
Lot of thanks for your help! 
 A: Define
$$
            f_n(t) = n\int_{0}^{t}\chi_{[1/2-1/2n,1/2+1/2n]}(u)du,
$$
where $\chi$ is the characteristic function of the given interval. $\{ f_n \}_{n=1}^{\infty}$ is a sequence of continuous functions, and it is a Cauchy sequence because $|f_n-f_{n+k}| \le 1$ and the difference function $f_n-f_{n+k}$ is non-zero only in the interval $[1/2-1/2n,1/2+1/2n]$. This gives the crude estimate
$$
            \|f_n -f_{n+k} \|^2 \le \frac{1}{n}.
$$
And, using the same type of estimate, it is clear that
$$
            \lim_n\int_{0}^{1}|f_n(t)-\chi_{[1/2,1]}(t)|^2dt = 0.
$$
If a subsequence $\{f_{n_m}\}_{m=1}^{\infty}$ converges to some $g\in C[0,1]$ in the norm of $L^2[0,1]$, then
$$
        \int_{0}^{1}|g(t)-\chi_{[1/2,1]}(t)|^2dt \\
   \le \int_{0}^{1}\{|g(t)-f_{n_{m}}|+|f_{n_{m}}(t)-\chi_{[1/2,1]}(t)|\}^2dt \\
   \le 2\int_{0}^{1}|g(t)-f_{n_{m}}(t)|^2+|f_{n_{m}}(t)-\chi_{[1/2,1]}(t)|^2dt.
$$
The far right side converges to $0$ as $m\rightarrow\infty$ and, hence,
$$
         \int_{0}^{1}|g(t)-\chi_{[1/2,1]}(t)|^2dt =0.
$$
Therefore,
\begin{align}
    0 & \le \int_{0}^{x}|g(t)-\chi_{[1/2,1]}(t)|^2dt \\
   & \le \int_{0}^{x}|g-\chi_{[1/2,1]}|^2dt+\int_{x}^{1}|g-\chi_{[1/2,1]}|^2dt \\
   & = \int_{0}^{1}|g-\chi_{[1/2,1]}|^2dt = 0\;\;\; 0 \le x \le 1,
\end{align}
which forces
$$
       \int_{0}^{x}|g-\chi_{[1/2,1]}|^2dt =0,\;\;\; 0 \le x \le 1.
$$
The integrand is continuous except possibly at $t=1/2$. Therefore, by the Fundamental Theorem of Calculus, for $x \in [0,1/2)\cup(1/2,1]$,
$$
      |g(x)-\chi_{[1/2,1]}(x)|^2= \frac{d}{dx}\int_{0}^{x}|g-\chi_{[1/2,1]}|^2dt = 0.
$$
So $g(x)=\chi_{[1/2,1]}(x)$ except possibly at $t=1/2$. So there is no way for $g$ to be continuous on $[0,1]$, regardless of the value of $g(1/2)$ because the left- and right-hand limits of $g$ at $1/2$ are not equal. This contradiction proves $\{ f_n \}$ has no convergent subsequence in your space.
