Pointwise convergence not enough to show incompleteness of continous functions from $[0,1]$ under the $L^2$ norm. If we want to show the space of continuous functions from $[0,1]$ under the $L^2$ 
norm $(f,g) = \int_0^1 f \bar g$ is not complete, then we have to find a Cauchy sequence in this space which does not converge in $C[0,1]$. If we take the sequence to be:
$f_n (x) = \begin{cases}
0 &\ \text{if}\ x\in [0, \frac{1}{2}),\\
n(x-\frac{1}{2}) &\ \text{if}\ x\in [\frac{1}{2}, \frac{1}{2}+\frac{1}{n})\\
1 &\ \text{if}\ x\in [\frac{1}{2}+\frac{1}{n},1].
\end{cases}$
Then it is not hard to show this sequence is Cauchy in $C[0,1]$. Next we need to show this sequence does not converge in this space. Intuitively as $n$ approaches infinity this sequence will point wise converge to 
$f(x) = \begin{cases}
0 &\ \text{if}\ x\in [0, \frac{1}{2}),\\ 
1 &\ \text{if}\ x\in [\frac{1}{2}, 1].
\end{cases}$ 
But $f(x$) is not in the space of continuous functions from $[0, 1]$. But technically this wouldn't be enough for a formal proof since we still haven't showed $f_n(x)$ still cannot converge to a function in the space. My approach to this proof would be by contradiction, but I want help making the technical proof rigorous.
 A: Suppose that it did converge to some function, $f \in C[0,1]$ in the $L^2$ norm. Then the integrals $$\int_0^1 |f - f_n|^2$$ go to zero, and so they must go to zero on every subinterval. If $x < \frac{1}{2}$, we can show that $f(x) = 0$, for if not, then by continuity we could choose a small interval about $x$ to make the integral above stay above some positive constant. A similar argument shows $f(x) = 1$ for $x > \frac{1}{2}$, and no continuous function can satisfy this.
A: You can consider the space $L^2([0,1])$ and show first that $f \in L^2([0,1])$ is the unique limit of $f_n$. Assume now that there is a continuous $f^*$ that represents $f$. By the same reasoning, that Zach L. applied, $f^*$ has to satisfy $f(x) = 0$, $x < \frac{1}{2}$, $f(x) = 1$, $x > \frac{1}{2}$. By continuity $0 = \lim_{x \rightarrow \frac{1}{2}^+} f(x) = \lim_{x \rightarrow \frac{1}{2}^-} f(x) = 1$, that is a contradiction. Hence $f^*$ cannot exist.
A: Hint: By Fatou's lemma, for any measurable function $f:[0,1]\to\mathbb R$,
\begin{eqnarray}
&&\int_0^{\frac{1}{2}}|f(x)|^2dx+\int_{\frac{1}{2}}^1|f(x)-1|^2dx \\
&=& \int_0^1\liminf_{n\to\infty}|f_n(x)-f(x)|^2dx \\
&\le& \liminf_{n\to\infty}\int_0^1|f_n(x)-f(x)|^2dx.
\end{eqnarray}
