# Show non-compactness of multiplication operator on $C[0,1]$

Show that the multiplication operator $$(Ax)(t):=(t+1)x(t)$$ in the Banachspace $C[0,1]$ is not compact.

Again I am struggling with compactness, it is always difficult to me to decide which criterion of compactness is the best to use in a special case.

1.) Should I try to show that $$\overline{\left\{Ax:x\in C[0,1], \lVert x\rVert_{\infty}\leq 1\right\}}$$ is not compact in $C[0,1]$?

2.) Or should I try fo find a bounded sequence $(x_n)\subset C[0,1]$ so that $(Tx_n)$ does not have a convergent subsequence?

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Most of the times the second approach will be easier. In this case, you can take $x_n(t)=t^n$.
Thanks for the example. I see that then $(x_n)$ is a bounded sequences, because $\lVert x_n\rVert_{\infty}=1$. But why does then $(Ax_n)$ not have a convergent subsequence? It is $\lVert Ax_n-Ax_m\rVert_{\infty}=\sup\limits_{t\in [0,1]}\lvert (t+1)t^n-(t+1)t^m\rvert=\sup\limits_{t\in [0,1]}\lvert (t+1)(t^n-t^m)\rvert$. I do not see that this is $\geq$ some limit, so that one cannot have a Cauchysequence. – math12 Feb 7 '13 at 12:07
$x_n(t)=(1+t)t^n$ converges pointwise to the function $x(t)=0$ if $0\le t<1$, $x(1)=2$. Since $x$ is not continuous, the convergence is not uniform. The same is true of any subsequence. – Julián Aguirre Feb 7 '13 at 13:22