$C[0,1]$ is incomplete with the norm $\|f\|_{2}=\sqrt{\int_{0}^{1}|f(x)|^2}$?

Question

Could any one give me an explicit example of a sequence which shows that $C[0,1]$ is incomplete with the norm $\|f\|_{2}=\sqrt{\int_{0}^{1}|f(x)|^2}$? Thank you for the help

Answer 1

Define $\{f_n\}_{n=2}^\infty$ as follows:

$f_n(x) = \Bigg\{ \begin{eqnarray} & 0 &, 0 \leq x < 1/2 \\ & n(x-1/2) & , 1/2 \leq x < 1/2 + 1/n \\ & 1 &, 1/2+1/n\leq x \leq 1 \end{eqnarray} \ \ $.

The function $f_n$ is in the interior points of all subintervals of the definition of domain a polynomial and hence continuous. At $x = \frac{1}{2}$ we have \begin{equation} \lim_{t \rightarrow \frac{1}{2}^-} f_n(t) = 0 = \lim_{t \rightarrow \frac{1}{2}^+} n(t-\frac{1}{2}) = \lim_{t \rightarrow \frac{1}{2}^+} f_n(t) \ \ . \end{equation} Hence $f_n$ is continuous at $x = \frac{1}{2}$. At $x = \frac{1}{2}+\frac{1}{n}$ we have \begin{equation} \lim_{t \rightarrow (\frac{1}{2}+\frac{1}{n})^-} f_n(t) = \lim_{t \rightarrow (\frac{1}{2}+\frac{1}{n})^-} n(t-\frac{1}{2}) = 1 = \lim_{t \rightarrow (\frac{1}{2}+\frac{1}{n})^+} f_n(t) \ \ . \end{equation} Hence $f_n$ is continuous at $x = \frac{1}{2}+\frac{1}{n}$. Hence $f_n \in C[0,1]$. Then we show that it is a Cauchy-sequence in $L^2$-norm. Note first that $f_n([0,1]) = [0,1]$. We have \begin{eqnarray} ||f_n-f_m||_2^2 & = & \int_0^1 |f_n(t)-f_m(t)|^2 dt = \int_0^\frac{1}{2} |f_n(t)-f_m(t)|^2 dt \\ & & + \int_\frac{1}{2}^{\frac{1}{2}+\frac{1}{\min\{n,m\}}} |f_n(t)-f_m(t)|^2 dt + \int_{\frac{1}{2}+\frac{1}{\min\{n,m\}}}^1 |f_n(t)-f_m(t)|^2 dt \\ & = & \int_\frac{1}{2}^{\frac{1}{2}+\frac{1}{\min\{n,m\}}} |f_n(t)-f_m(t)|^2 dt \\ & \leq & \int_\frac{1}{2}^{\frac{1}{2}+\frac{1}{\min\{n,m\}}} (|f_n(t)|+|f_m(t)|)^2 dt \leq \frac{4}{\min\{n,m\}} \rightarrow 0 \end{eqnarray} as $n,m \rightarrow \infty$. This shows that $f_n$ is a Cauchy-sequence.

Make now the claim that $L^2(C[0,1])$ is incomplete. Proof by contradiction: Assume that $L^2(C[0,1])$ is complete. Then by definition there is an element $f\in C[0,1]$ s.t. $f = \lim_{n \rightarrow \infty} f_n$ in $L^2$-norm. We show by contradiction that $f(x)=0$ for $0 \leq x < 1/2$ and $1$ for $1/2 < x \leq 1$: 1) Assume that there is $x \in [0,1/2)$ s.t. $f(x) \neq 0$. Then there is a neighborhood $\mathcal{U}_\delta(x) \subset [0,1/2]$ s.t. $|f(x') - f(x)| \leq \frac{1}{2}|f(x)|$ for every $x' \in \mathcal{U}_\delta(x)$. We have \begin{eqnarray} |f_n(x')-f(x')| & = & |0-f(x')| = |f(x)+f(x')-f(x)| \\ & \geq & ||f(x)|-|f(x')-f(x)|| \\ & \geq & |f(x)|-|f(x')-f(x)| \\ & \geq & |f(x)|-\frac{1}{2}|f(x)|= \frac{1}{2}|f(x)| \ \ . \end{eqnarray} Hence \begin{eqnarray} ||f_n-f||_2^2 & = & \int_0^1 |f_n(t)-f(t)|^2 dt \geq \int_{\mathcal{U}_\delta(x)} |f_n(t)-f(t)|^2 dt \\ & \geq & \int_{\mathcal{U}_\delta(x)} (\frac{1}{2}|f(x)|)^2 dt = \int_{\mathcal{U}_\delta(x)} \frac{1}{4}|f(x)|^2 dt  \\ & \geq & \int_x^{x+\delta} \frac{1}{4}|f(x)|^2 dt = \frac{\delta}{4}|f(x)|^2 \\ ||f_n-f||_2 & \geq & \frac{\sqrt{\delta}}{2}|f(x)| \end{eqnarray} for every $n\in\{i\}_{i=2}^\infty$. This contradicts $f = \lim_{n \rightarrow \infty} f_n$. Hence $f(x) = 0$ for every $x \in [0,1/2)$. 2) Assume that there is $x \in (1/2,1]$ s.t. $f(x) \neq 1$. Set $N \in \mathbb{N}$ s.t. $1/2+1/N<x$. There is a neighborhood $\mathcal{U}_\delta(x) \subset [1/2+1/N,1]$ s.t. $|f(x')-f(x)| \leq\frac{1}{2}|1-f(x)|$ for every $x \in \mathcal{U}_\delta(x)$. Then for $n > N$ we have \begin{eqnarray} |f_n(x')-f(x')| & = & |1-f(x')| = |1-f(x)+f(x)-f(x')| \\ & \geq & ||1-f(x)|-|f(x)-f(x')|| \\ & = & ||1-f(x)|-|f(x')-f(x)|| \\ & \geq & |1-f(x)|-|f(x')-f(x)| \\ & \geq & |1-f(x)|-\frac{1}{2}|1-f(x)| = \frac{1}{2}|1-f(x)| \ \ . \end{eqnarray} Hence \begin{eqnarray} ||f_n-f||_2^2 & = & \int_0^1 |f_n(t)-f(t)|^2 dt \geq \int_{\mathcal{U}_\delta(x)} |f_n(t)-f(t)|^2 dt \\ & \geq & \int_{\mathcal{U}_\delta(x)} (\frac{1}{2}|1-f(x)|)^2 dt \geq \int_{x-\delta}^x \frac{1}{4}|1-f(x)|^2 dt \\ & = & \frac{\delta}{4} |1-f(x)|^2 \\ ||f_n-f||_2 & \geq & \frac{\sqrt{\delta}}{2} |1-f(x)| \ \ . \end{eqnarray} This contradicts $f = \lim_{n \rightarrow \infty} f_n$. Hence $f(x) = 1$ for every $x \in (1/2,1]$. By continuity of $f$ $0 = \lim_{t \rightarrow \frac{1}{2}^-} f(t) = \lim_{t \rightarrow \frac{1}{2}^+} f(t) = 1$, that is a contradiction. Hence $L^2([0,1])$ is incomplete.