Let $(E,\|\cdot \|_1)$ be the normed space of continuous functions from $[0,1]$ to $\mathbb R$. Let's consider $F$ the subset of $E$ of functions verifying $f(0)=0$. I want to show that $F$ is dense in $E$. My idea is to take a function $f\in E$ and try to come up with a sequence of functions $(f_n)$ in $F$ such that $(f_n)$ converge to $f$; that is we need :
$f_n$ is continuous on $[0,1]$ for each $n$,
$f_n(0)=0$ for each $n$ and
$\displaystyle\lim_{n\to \infty} \int_0^1 \left|f_n(x)-f(x)\right| dx = 0$
I guess that $f_n$ will be $f$ on some parts depending on $n$ of $[0,1]$ and zero otherwise but I could not write this correctly.
Let $f_n:[0,1]\rightarrow\mathbb R$ be given by $f_n(x)=f(1/n)n x$ for $x\in[0,1/n]$ and $f_n(x)=f(x)$ otherwise. The left and right limits at $1/n$ exist and coincide, otherwise the function is continuous. Finally, we have $\int|f_n(x)-f(x)|\mathrm dx=\int_0^{1/n}|f_n(x)-f(x)|\mathrm dx\le\int_0^{1/n}|f_n(x)|\mathrm{d}x+\int_0^{1/n}|f(x)|\mathrm{d}x\le 2\|f\|_\infty/n$, which also shows that $f_n$ is integrable. Of course, in the last inequality it would be sufficient to take $\max_{0\le x\le 1/n}|f(x)|$, which is taken since we consider a continuous function on a compact set.