# Showing closure in the $\|\cdot\|_1$ norm

Let $X=C[0,1]$ and $W=\{f\in X\mid f(0)=0\}$. What is the closure of $W$ wrt the 1-norm $\|\cdot\|_1$.

My solution is as follows:

The closure is the whole space $X$. To see this take any function $f(x)\in X$, assume with out loss of generality that $f$ is in the upper right plane. We will construct a sequence that converges to this $f(x)$.

Consider the sequence $f_\epsilon(x)= \begin{cases} f(x), & \text{if }x\in [0,1]-(0,\epsilon), \\ \frac{f(\epsilon)}{\epsilon}x, & \text{if }x\in(0,\epsilon). \end{cases}$

Then if we let $\epsilon\rightarrow 0$ $f_{\epsilon}\rightarrow f$ and so we have that the limit points are the whole of $X$.

Is this correct?

Thanks for any help

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Looks good. Think you mean "if we let $\epsilon \to 0$, $f_\epsilon \to f$ and ...", though, no? –  fgp Oct 11 '12 at 18:25
You need to have the $f_n \in W$. Why not try something like $f_n(x) = f(x)$ for $x \in [\frac{1}{n},1]$ and $f_n(x) = x f(\frac{1}{n})$ otherwise. This way you have $f_n \in W$. Then you can show that $\|f - f_n\|_1 \to 0$. –  copper.hat Oct 11 '12 at 18:34
@fgp yeah thanks I've made the edit –  hmmmm Oct 11 '12 at 18:38
@copper.hat I forgot to put an x in somewhere byt my $f_\epsilon$ is in W now right? –  hmmmm Oct 11 '12 at 18:40
Yes, that is good. (I forgot a $\frac{1}{n}$ in my formula.) –  copper.hat Oct 11 '12 at 18:42

As copper.hat also mentioned, perhaps $\epsilon=1/n$ is easier to argue, and perhaps one more sentence about why $f_\epsilon\to f$ in $L_1$-norm, but that's all.