Find the closure of $C=\{f\in C([0,1]):f(0)=0\}$

Question

Let $C([0,1])$ be the space of all real valued continuous functions $f:[0,1]\to \mathbb{R}$. Take the norm $$||f||=\left(\int_0^1 |f(x)|^2\right)^{1/2}$$ and the subspace $$C=\{f\in C([0,1]):f(0)=0\}.$$ Find the closure of $C$.

First, I showed that $C$ is not closed, so it cannot be its own closure. To do this, let's take the function $$f_n(x)=\left\{ \begin{array}{ll} nx & \text{if } x\leq \tfrac{1}{n} \\ 1 & \text{if } x\geq \tfrac{1}{n} \end{array} \right.$$ Clearly, $f_n(x)\in C$ for any $n\in \mathbb{N}$. We'll prove that $$f:=\lim_{n\to \infty} f_n(x)=\left\{ \begin{array}{ll} 0 & \text{if } x=0 \\ 1 & \text{if } x\neq 0 \end{array} \right.$$ To see this, let $\varepsilon>0$ and choose $N\in \mathbb{N}$ such that $\frac{1}{\sqrt{3N}}\leq \varepsilon$. Then, for any $n\geq N$ we have that $$\int_0^1|f_n(x)-f(x)|^2dx=\int_0^{1/n}|f_n(x)-f(x)|^2dx+\int_{1/n}^1|f_n(x)-f(x)|^2dx$$ $$\int_0^{1/n}|nx-1|^2dx+\int_{1/n}^1|1-1|^2dx=\int_0^{1/n}(nx-1)^2dx=\frac{1}{3n}\leq \frac{1}{3N}\leq \varepsilon.$$

Hence, we have constructed a sequence in $C$ which does not converge in $C$ since $f$ is not continuous, so $C$ cannot be closed.

I don't know how to approach the question of finding the closure of $C$. Any help would be appreciated.

Answer 1

Lemma.

Let $X$ be a normed space and $f : X \to \mathbb{C}$ an unbounded functional on $X$. Then $\ker f$ is dense in $X$.

Proof. Pick a sequence $(x_n)_n$ in $X$ such that $\|x_n\| = 1$ and $|f(x_n)| \ge n, \forall n \in \mathbb{N}$. Let $x \in X$ be arbitrary. Check that the sequence $\left(x-\frac{x_n}{f(x_n)}f(x)\right)_n$ lies in $\ker f$ and converges to $x$.

Now, the functional $\phi : C[0,1] \to \mathbb{C}$ given by $\phi(f) = f(0)$ is unbounded. Namely, consider the functions $$g_n(t) = \begin{cases} \sqrt{2n - 2n^2t}, &\text{if } t \in \left[0,\frac1n\right]\\ 0, &\text{if } t \in \left[\frac1n,1\right]\end{cases}$$

We have $\|g_n\|_2 = 1$ but $\phi(g_n) = g_n(0) = \sqrt{2n}$.

Hence $C = \ker \phi$ is dense in $C[0,1]$, therefore $\overline{C} = C[0,1]$.

Answer 2

The closure of $C$ in $V=(C([0,1]),\|\cdot\|_2)$ is all of $V$. For given $f\in V$, let $f_n\in C$ be obtained from $f$ by $$ f_n(x)=\begin{cases}xf(1/n) \text{ if } x\in[0,1/n]\\ f(x) \text{ otherwise.}\end{cases}$$ It's clear that $\|f_n-f\|_2\to0$, so $f$ is in the closure of $C$.