I'm working through the functional analysis book by Milman, Eidelman, and Tsolomitis, and I have a question. The book states a lemma that I'm a bit confused about:

A sequence of operators $T_n\in L(X, Y)$ (here, $X$ and $Y$ are Banach Spaces) converges strongly to an operator $T\in L(X,Y)$ if and only if

(i) the sequence $\{T_n x\}$ converges for any $x$ in a dense subset $M\subset X$;

(ii) there exists $C>0$ such that $\| T_n\|\leq C$.

My question is whether or not $M$ has to be a linear subspace. I think it does. The reason I think so is that the proof of the theorem says that we first define an operator $T_0$ by $$ T_0 x:= \lim_{n\to \infty} T_n x. $$ By the assumption that $T_n x$ converges on $M$, we have that the domain of an operator is $M$. Hence, $M$ must be a linear space. The proof then goes on to define an extension of $T_0$ by $$ Ty:= \lim_{n\to \infty} T_0 x_n, $$ where $x_n\to y$. Here we are using the density of $M$ and the boundedness of $T_0$ which is inherited from $\{T_n\}$. Even here though, to prove uniqueness of this limit under any sequence converging to $y$ we need the fact that $M$ is a subspace. The reason is that, if $\{z_n\}$ is any other sequence in $M$ converging to $y$, we have to consider the expression $$ \|T_0 x_n -T_0 z_n\|=\|T_0(x_n-z_n)\|\leq \|T_0\|\cdot \|x_n-z_n\|\to 0$$ since $\{x_n\}$ and $\{z_n\}$ both converge to $y$. We must have that the vector $x_n-z_n$ is in the domain of $T_0$ (i.e. $M$).

Anyways, I'm pretty sure that our dense subset $M$ must be a dense subspace for this to work. Can someone please tell me if this is correct. I'm 98% sure this is a typo in the statement of the lemma.

  • $\begingroup$ Yes, if $T_nx$ converges for every $x\in M$, you can set $T_0x:=\lim T_nx$ for $x\in M$. But this will never force anything about $M$. At least not that it be a subspace. For instance, consider the dense subset $ M=\mathbb{Q}$ of the Banach space $\mathbb{R}$ equipped with the usual norm. Then the sequence $T_n=Id$ converges pointwise to $Id$ on $\mathbb{Q}$, and is even uniformly bounded by $1$. This does not make $\mathbb{Q}$ a real linear subspace of $\mathbb{R}$. But anyway, you only need $M$ to be dense here. $\endgroup$
    – Julien
    Jul 11, 2013 at 7:29

2 Answers 2


Of course, $L(X,Y)$ means bounded linear operators.

That's a standard $\frac{\epsilon}{3}$ argument which does not require the linearity of $M$. Just that it be dense.

1- Assume (i) and (ii) are fulfilled.

Now take any $x\in X$. Let us prove that the sequence $\{T_nx\}$ is Cauchy. So let us take $\epsilon>0$. By density of $M$, there exists $y\in M$ such that $\|x-y\|\leq \frac{\epsilon}{3C}$. Then $$ \|T_nx-T_mx\|\leq \|T_nx-T_ny\|+ \|T_ny-T_my\|+\|T_my-T_mx\| $$ $$ \leq \|T_n\|\|x-y\|+\|T_ny-T_my\|+\|T_m\|\|x-y\| $$ $$ \leq C\frac{\epsilon}{3C}+\|T_ny-T_my\|+C\frac{\epsilon}{3C} $$ $$ =\frac{2\epsilon}{3}+\|T_ny-T_my\|. $$ Since $\{T_ny\}$ converges, we can find $N$ such that $\|T_ny-T_my\|\leq \frac{\epsilon}{3}$ for all $n,m\geq N$. Whence $\|T_nx-T_mx\|\leq \epsilon$.

Since $Y$ is complete, it follows that $\{T_nx\}$ converges to some $Tx$ for every $x\in X$. Since the pointwise limit of a sequence of linear operators is linear, $T$ is linear. And since $\|T_nx \|\leq C\|x\|$ for every $x$, we get $\|Tx\|\leq C\|x\|$ as well at the limit. Whence $T$ is bounded. This direction requires that $Y$ be complete, but not $X$.

2- Conversely, it $\{T_n\}$ converges strongly, then it must be uniformly bounded by the uniform boundedness principle (and its limit must be bounded). So (i) and (ii) are satisfied by $M=X$. This direction requires that $X$ be complete. But not $Y$.

  • $\begingroup$ Thanks for your response. It cleared up my confusion. $\endgroup$ Jul 11, 2013 at 14:55
  • $\begingroup$ @AlexLapanowski You're welcome. Apparently your confusion came from the approach suggested by the book... $\endgroup$
    – Julien
    Jul 11, 2013 at 14:58

No, it need not be assumed that $M$ is a linear subspace. Without making that assumption, one can reduce to the case where $M$ is a linear subspace by noting that $\{x\in X:(T_nx)\text{ converges}\}$ is a linear subspace of $X$ containing $M$, and therefore it is a dense linear subspace of $X$. I think that making this reduction makes the proof clearer, but it is not necessary to put it in the statement of the lemma.

  • $\begingroup$ Actually, I can't see where the linearity of $M$ helps in the proof. Do you have another argument than mine in mind? $\endgroup$
    – Julien
    Jul 11, 2013 at 7:10
  • $\begingroup$ @julien: I was referring to the proof that Alex had in mind (to the extent that it was sketched above). He wants $T_0$ to be defined on a linear subspace at a particular step, which would be automatic if we just assume WLOG that $M$ is linear. I like your argument. $\endgroup$ Jul 11, 2013 at 7:13
  • $\begingroup$ Hmm. I stopped at "we have that the domain is $M$. Hence $M$ must be linear". Which in this case is exactly like saying the real identity function is well-defined on $\mathbb{Z}+e\mathbb{Z}+\pi$. Hence $\mathbb{Z}+e\mathbb{Z}+\pi$ is a subgroup of $\mathbb{R}$. $\endgroup$
    – Julien
    Jul 11, 2013 at 7:21
  • $\begingroup$ A much better reason for the confusion came later. I don't have the book, but I get the impression that linearity of $M$, or extension of $T_0$ to the span of $M$, might have been implicitly assumed. $\endgroup$ Jul 11, 2013 at 7:24

You must log in to answer this question.

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