# Show strong operator convergence

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.

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.

• @postmortes I initially used the approach I mentioned to show strong convergence and I did try what you said but I can't seem to figure out how $T_nx$ converges to anything.
– rev
May 4, 2021 at 14:33
• @postmortes Yes but I thought that strong convergence, in concrete terms, meant $||T_nx - Tx|| \to 0$
– rev
May 4, 2021 at 14:37
• I see, thank you I really appreciate your patience
– rev
May 4, 2021 at 14:40
• @postmortes I added an answer with basically what is being discussed in the comments and a small explanation to why the convergence occurs:) May 4, 2021 at 14:54

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 $$0$$.

• Thank you for the explanation. But I was wondering if you could also tell me whether if I was successful at showing that the sequence isn't uniformly convergent or not?
– rev
May 4, 2021 at 14:57
• @rev Sure. I'm afraid what you present is not correct. You say that $\|T_n\|=c$ and that since we cannot say $c=0$ this does not show uniform convergence. First, if $c=0$ then $T_n=0$! so what's the point? maybe you wanted to say that $\|T_n\|=c_n$ and that "since we cannot say $c_n\to0$ this does not show uniform convergence"? Again, it would be wrong though for two reasons: a) if we cannot say this, then there is still a chance that it is true but we failed to prove it, so we simply don't know based on that reason. and b) even if $T_n$ was convergent uniformly, it would converge uniformy, May 4, 2021 at 15:02
• @rev it would converge uniformly to its strong limit, i.e. to the operator $I$. So you would have to check whether $\|T_n-I\|$ converges to $0$ or not. If it converges to $0$, then $T_n\to I$ "uniformly", i.e. in norm. If not, then it doesn't. May 4, 2021 at 15:03
• I see, thank you again for the detailed explanation
– rev
May 4, 2021 at 15:05
• @rev If you have trouble checking whether $\|T_n-I\|\to0$ or not, let me know and I will add the details to my answer, but I recommend trying it out, it will be a good exercise after this conversation:) May 4, 2021 at 15:11