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