# Convergence in operator norm of $C[0, 1]$

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!

## 2 Answers

$$t^{n}(1-t) \to 0$$ 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.

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

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