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Does the sequence $A_{n}x(t)=t^{n}(1-t)x(t)$ in $C[0, 1]$ converge in the operator norm? Thanks in advance!

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$t^{n}(1-t) \to 0$ in the supremum norm. Since $\|A_nx\| \leq \|x\| sup \{t^{n}(1-t):0\leq t \leq 1\}$ it follows that $\|A_n\| \to 0$. To show that $t^{n}(1-t) \to 0$ in the supremum norm let $\epsilon >0$. For $1-\epsilon \leq t \leq 1$ we have $t^{n}(1-t) \leq t^{n}\epsilon \leq \epsilon$. For $0\leq t \leq 1-\epsilon$ we have $t^{n}(1-t) \leq (1-\epsilon) ^{n} <\epsilon$ for $n$ sufficiently large.

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As for the norm convergence, since strong convergence is answered already, note that $$ \begin{array}{rl}\| A_n - A_m\| &= \sup_{\| x \| \leq 1}\| A_nx - A_mx\| = \sup_{\| x \| \leq 1}\| t^n(1-t)x(t) - t^m(1-t)x(t)\| \\ &= \sup_{\| x \| \leq 1}\|x(t)\|\| (1-t)(t^n - t^m)\|_{\infty} \end{array} $$

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  • $\begingroup$ The inequality in Kavi's answer shows directly that $\|A_n\|\leq\frac1{n+1}$. $\endgroup$ – Martin Argerami Oct 27 '18 at 14:59

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