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.

  • $\begingroup$ @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. $\endgroup$
    – rev
    May 4, 2021 at 14:33
  • $\begingroup$ @postmortes Yes but I thought that strong convergence, in concrete terms, meant $||T_nx - Tx|| \to 0$ $\endgroup$
    – rev
    May 4, 2021 at 14:37
  • $\begingroup$ I see, thank you I really appreciate your patience $\endgroup$
    – rev
    May 4, 2021 at 14:40
  • $\begingroup$ @postmortes I added an answer with basically what is being discussed in the comments and a small explanation to why the convergence occurs:) $\endgroup$ May 4, 2021 at 14:54

1 Answer 1


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.


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 $0$.

  • $\begingroup$ 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? $\endgroup$
    – rev
    May 4, 2021 at 14:57
  • $\begingroup$ @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, $\endgroup$ May 4, 2021 at 15:02
  • $\begingroup$ @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. $\endgroup$ May 4, 2021 at 15:03
  • $\begingroup$ I see, thank you again for the detailed explanation $\endgroup$
    – rev
    May 4, 2021 at 15:05
  • $\begingroup$ @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:) $\endgroup$ May 4, 2021 at 15:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .