I have a problem in which I need to prove that a sequence of operators is strongly convergent and that it isn't uniformly convergent. The operators are defined like so: $ T_j: L^1 \to L^∞ $
And the sequence goes on as such:
$$
T_1(x) = (x_1, x_1, x_1, x_1,…)
$$
$$
T_2(x) = (x_1, x_2, x_2, x_2,…)
$$
$$
T_3(x) = (x_1, x_2, x_3, x_3,…)
$$
$$
T_n(x) = (x_1, x_2, x_3, …, x_n, x_n,…)
$$
For the first part of the problem I couldn't really come up with anything because I'm new to functional analysis but for the second part in which I tried to prove that the sequence isn't uniformly convergent, I tried this:
$$
||T_nx|| = ||(x_1, x_2, x_3, …, x_n, x_n,…)||\\ = ||(x_1, x_2, x_3, …, 0, 0,…) + (0, 0, 0, …, x_n, x_n,…)||
$$
$$
||T_nx|| \le \sup\limits_{i}||x_i|| + |x_n| = c\\ ||T_n|| = \sup \frac{||T_nx||}{||x||} = \sup\limits_{||x|| = 1}||T_nx|| = c
$$
Since we cannot say that $c$ ($\ge0$) equals to zero, $||T_n – 0|| = ||T_n||$ does not converge uniformly to zero.
First of all I would really like to know if my solution to the second part is true or not. Also, if you can give me any ideas as to how I can show strong convergence or provide solutions, I would really appreciate it.
Thank you in advance
Edit 1: As @postmortes suggested, to show strong operator convergence, looking at $\lim \limits_{n \to ∞}||T_nx – x||$ we can see that $T_nx$ converges to zero since $$T_nx – x = (0,0,0,…,0,x_n – x_{n+1}, x_n – x_{n+2},…)$$ and as $n\to∞$ this is going to be zero so $T_nx\to x$ (Strong convergence means that $||T_nx – Tx|| \to 0$ and here that $T$ operator is the identity operator $I$ where $Ix = x$)
Edit 2: As @JustDroppedIn stated, my way of showing that the sequence does not converge uniformly is wrong.
Best Answer
Note that if $x=(x_n)\in\ell^1$, then $x\in\ell^\infty$ as well. Now
$$\|T_nx-x\|_{\ell^\infty}=\|(0,0,,\dots,0,x_n-x_{n+1},x_n-x_{n+2},x_n-x_{n+3},\dots)\|_\infty=\sup_{k\geq1}|x_n-x_{n+k}|$$ But since $x\in\ell^1$ we have that $\{x_n\}$ is a Cauchy sequence (since it converges (to $0$)). So if $\varepsilon>0$ there exists $N\geq1$ so that $|x_n-x_m|<\varepsilon$. Then, for $n\geq N$ we have that $$\|T_nx-x\|_\infty=\sup_{k\geq1}|x_n-x_{n+k}|\leq\varepsilon$$ which shows that $T_nx\to x$ in $\ell^\infty$. In other words, since this is true for all $x$, we have that $T_n\to I$ strongly, where $I:\ell^\infty\to\ell^\infty$ is the identity operator.
comment: Note that the same would be true if we "enlarged" the domain of each $T_n$ and used the space $c$ of convergent sequences with supremum norm.
Edit
Here are some details about checking whether $T_n$ converges in norm, as OP seems to have trouble with this part.
First, if $T_n$ converges in norm to something, it has to be $I$, because if $T_n$ converges in norm to $S$, then $T_n$ converges strongly to $S$ and therefore $S=I$ (strong operator limits are unique).
First let's compute the norm of $T_n$, but this is not needed to check whether $T_n\to I$ in norm; we should be computing $\|T_n-I\|$ for this, but let's start with that. Note that $n$ is fixed now! We have that $$\|T_n\|=\sup_{\|x\|_{\ell^1}=1}\|T_nx\|_{\ell^\infty}=\sup_{\|x\|=1}\|(x_1,\dots,x_n,x_n,x_n,\dots)\|_{\ell^\infty}=\sup_{\|x\|=1}\bigg\{\max_{1\leq j\leq n}|x_j|\bigg\}\leq\sup_{\|x\|=1}\bigg\{\sum_{j=1}^\infty|x_j|\bigg\}=1 $$ So $\|T_n\|\leq1$. On the other hand, if $x=(1,0,0,\dots)$ then $\|x\|_{\ell^1}=1$ and $T_nx=(1,0,0,\dots,)$ and $\|T_nx\|_{\ell^\infty}=1$, so $$\|T_n\|=\sup_{\|y\|_{\ell^1}}\|T_n(y)\|\geq\|T_n(x)\|=1$$ and this shows that $\|T_n\|=1$ for all $n$.
Now let's compute in the same fashion the norm of $T_n-I$, again $n$ is fixed. We do not really need to do an exact calculation, we only need to know if $\|T_n-I\|$ converges to $0$ or not. By our earlier computations before the edit, $$\|T_n-I\|=\sup_{\|x\|=1}\|T_nx-x\|=\sup_{\|x\|=1}\sup_{k\geq1}|x_n-x_{n+k}|$$ Now consider $x=(0,\dots,0,0,1,0,\dots)$ where $1$ appears in the $n+1$ position. Then, $\sup_{k\geq1}|x_n-x_{n+k}|=sup_{k\geq1}|x_{n+k}|=|x_{n+1}|=1$. Also note that $\|x\|_{\ell^1}=1$, so, $$\|T_n-I\|\geq\|T_nx-x\|_{\ell^\infty}=1$$ This shows that the sequence $\{\|T_n-I\|\}_{n=1}^\infty$ is ALWAYS greater than or equal to $1$, so it cannot converge to $0$.