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First of I just want to provide the definitions I have for the three concepts. Let $A_n: X \to Y$.

$A_n$is said to converge to $ A:(n\to \infty)$

$$\text{Uniformly : }\|A_n-A\|=0 \\ \text{Strongly : }\lim_{n \to \infty}A_n(x) = A(x) \\ \text{Weakly :} \lim_{n \to \infty}y^*(A_nx)=y^*(Ax) :\forall y^*\in X^*$$

Sometimes, I come across that strong convergence is $\lim_{n \to \infty}\|A_n(x)-A(x)\|=0$, which is logical.

Anyway: I am to question the convergence of the following:

$$a.) \ \ D_n : C[0,1]\to C[0,1]: D_n(x(t))=t^n(1-t)x(t),x \in C[0,1] \\ b.) \ \ E_n : L_2[0,1]\to L_2[0,1]: E_n(x(t))=\int_{0}^{1}t^nsx(s)ds,x \in L_2[0,1] \\ c.) \ \ F_n : \ell_2 \to \ell_2: F_n(x)=(\frac{x_1}{n^2},\frac{x_2}{n^2},\frac{x_3}{n^2},...),x=(x_1,x_2,...) \in \ell_2 \\$$

I also have the theorem that:

Let $\{A_n\}_{n=1}^{\infty}$ be a sequence $A_n:X \to Y$ ($X,Y$ are Banach spaces) such that $\|A_n\|\leq M, M>0.$ Let $E $ be a fundamental set in $X$.

If $\exists \lim_{n \to \infty}A_n(x), \forall x \in E\implies \exists > \lim_{n \to \infty}A_n(x),\forall x \in X$

* ------My solutions-------- *

I know that the fundamental set in $C[0,1]$ is the set $\{1,t,t^2,...\}$ and in $\ell_2: \{(1,0,..),(0,1,0,0,..),(0,0,1,0,..),...\}.$ How about in $L_2[0,1]?$ Do I have to take this route of the theorem. Because, for example for $a.$ I tried:

$\lim_{n \to \infty}A_n(x(t))=0;$(because $t \in (0,1)$) to I check the uniform by:

$$0 \leq \|A_n(x(t))\|_{C[0,1]} \leq \|x\| \lim_{n\to \infty}\max_{t\in[0,1]}(t^n(1-t)) \implies \\ 0 \leq \frac{\|A_n(x(t))\|_{C[0,1]}}{\|x\|} \leq \lim_{n\to \infty}\max_{t\in[0,1]}(t^n(1-t))=0 \implies$$ that this sequence converges uniformly to 0, also strongly and weakly because of inclusion relation of the three.

For the second: $\lim_{n\to \infty}A_n(x(t))= \begin{cases} 0 & \text{if $t \in [0,1)$} \\ \int_{0}^{1}sx(s)ds & \text{if $t = 1$.} \end{cases}$. I don't know how to work with this.

For the third I let the limit go to $\infty$ and question the convergence of the sequence to $(0,0,0,...)$. Is this valid what I am doing in your opinion? There is always the possibility that the sequences don't converge, like having a fixed distance in between two elements, but, what do you all think?

I have my exam in two days, I need to clear this up. Thanks for all input.

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The most important thing is to rewrite your sequences so that it is clear what is happening. For example once you see that $F_n$ is just $\mathbb1/n^2$ it is obvious that $F_n\to0$ uniformly.

What does $E_n$ do? $$E_n(x)\,(t)=t^n\int_0^1 s x(s)\,ds = t^n\,\langle \mathrm{id},x\rangle\tag{1}$$ So $$\|E_n(x)\|_2=\|t^n\|_2\,|\langle \mathrm{id},x\rangle≤\|t^n\|_2\,\|x\|_2\,\|\mathrm{id}\|_2=\frac{\|x\|_2}{2(n+1)}$$ and $\frac{\|E_n(x)\|}{\|x\|}≤\frac1{2(n+1)}$, thus $\|E_n\|=\sup_{x\neq0}\frac{\|E_n(x)\|}{\|x\|}≤\frac{1}{2(n+1)}$ and $E_n\to0$ uniformly as well.

The way to see it immediatly is to understand that equation $(1)$ is of the form $E_n= A_n \circ B$, with $B:L^2\to\Bbb C, x\mapsto \langle\mathrm{id},x\rangle$ and $A_n:\Bbb C\to L^2, z\mapsto (t\mapsto t^n\,z)$. It follows $\|E_n\|≤\|A_n\|\,\|B\|$, where $\|A_n\|\to0$.

You have seen correctly that $D_n$ converges uniformly also.

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  • $\begingroup$ Thank you so much, I did not see this. $\endgroup$ – Bocko Jun 13 '17 at 7:18

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