show that the following sequence of operators $T_n$ does not converge uniformly

Question

I am unable to show that the following sequence of operators $T_n : \ell^2 \to \ell^2$ does not converge uniformly:

1. $T_n(x)=(x_1,x_2,x_3,\ldots ,x_n,0,0\ldots )$
2. $T_n(x)=(x_{n+1},x_{n+2},x_{n+3},\ldots)$

1. $$\|(T_n-T_m)(x)\|=\left(\sum_{i=m+1}^n x_i^2\right)^{\frac{1}{2}}$$

I don't understand how to show that $\|T_n-T_m\|\to 0$ is not true.

2. $$\|(T_n-T_m)(x)\|=\left(\sum_{i=m+1}^n (x_i-x_{i+1})^2\right)^{\frac{1}{2}}$$

I don't understand how to show that $\|T_n-T_m\|\to 0$ is not true.

I need some help. Can anyone please do it?

Answer 1

For 1) If $\{e_n\}$ is the standard basis then $T_n(e_n)=e_n$ and $T_m(e_n)=0$ for $m <n$. Hence $\|T_n-T_m\|\geq 1$.

2) is similar and I leave it to you.

Answer 2

1. Clearly $T_n\to I$ pointwise so the only candidate for uniform convergence is also the identity map $I$. However, $$\|I - T_n\| \ge \|(I - T_n)(e_{n+1})\|_2 = 1$$ so $T_n \not\to I$ uniformly.

2. We have $$\|T_n(x)\|_2^2 = \sum_{k=n+1}^\infty |x_k|^2 \xrightarrow{n\to\infty} 0$$ so $T_n \to 0$ pointwise. Hence the only candidate for uniform convergence is also $0$. However, $$\|T_n\| \ge \|T_n(e_{n+1})\|_2 = 1$$ so $T_n \not\to 0$ uniformly.