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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

  • 1
    Yes, that is good. (I forgot a $\frac{1}{n}$ in my formula.)2012-10-11

1 Answers 1

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Yes. Perfect.

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.