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I tried to prove the following statement and would like to ask if it's okay like that.

$\underline{Statement}$:

Let [B,$||.||_B$] be a complete normed linear space, $T_n :B \rightarrow B$ a sequence of linear maps which are bounded uniformly in n. Let $D \subset B$ be a dense subset of B and $T_nx \rightarrow Tx$ if $x \in D$.
Then $T_nx \rightarrow Tx \ \ \forall x\in B $.

$\underline{Proof}:$

$(x_n)_n$ Cauchy sequence in B and B complete $\curvearrowright$ $x_n \rightarrow x \in B$, so that with $j>j_0$: $$ ||T_nx_j - T_nx||_B \leq ||T_n||~||x_j - x||_B < \frac{\epsilon}{3} \ \ (1) $$

$(x_n)_n$ Cauchy sequence in B and $j,k > k_0$, then: $$ ||T_nx_j - T_nx_k||_B \leq ||T_n||~||x_j - x_k||_B < \frac{\epsilon}{3} \ \ (2) $$

D dense in B $\curvearrowright \forall x \in B \ \exists (y_j) \subset D:~ \lim_{j \to \infty} y_j = x $;

with $y_j \in D$, $\epsilon$ fixed and $ n>n_0 \curvearrowright ||T_ny_j - Ty_j||_B < \frac{\epsilon}{3}$ so that $$ T_ny_j \rightarrow T_nx \ \ (3) $$

Pick any fixed $x \in B$, J = max{$j_0,k_0$}, then for $j,k > J$, $n > n_0$

$$ ||T_nx - Tx||_B \leq \underbrace{||T_nx-T_nx_j||_B}_{<\frac{\epsilon}{3},~(1)} + \underbrace{||T_nx_j - T_ny_j||_B}_{<\frac{\epsilon}{3},~(2)} + \underbrace{||T_ny_j - Tx||_B}_{<\frac{\epsilon}{3},~(3)} < \epsilon $$

JtSpKg
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    Your proof starts with “$(x_n)_n$ Cauchy sequence in $B$”. What do you mean? Is it a specific Cauchy sequence. Is what follows true for all Cauchy sequences? Besides, what does $\curvearrowright$ mean? – José Carlos Santos Aug 17 '17 at 14:10
  • (+1 for taking the time to write up the question decently, incl. formatting!) – StackTD Aug 17 '17 at 14:13
  • You should specify that $T$ is also a continuous linear operator. – John D Aug 17 '17 at 14:29
  • Hey, thanks for the fast answers, @JoséCarlosSantos I wanted to pick any Cauchy sequence $(x_n)_n \subset B$ , so the following should be true for all Cauchy sequences. By "$A \curvearrowright B$" I mean "A implies B". – JtSpKg Aug 18 '17 at 11:02
  • @Magnusseen I thought it would be clear from the context, since a linear operator $A$ between two normed linear spaces ist bounded iff it is continous – JtSpKg Aug 18 '17 at 11:09
  • @JtSpKg No. Of course continuity equals boundedness. What i meant was that $T$ is not defined in your statement. It just appears at the end of a limit. Either you say that $T: B \to B$ is an operator and we must prove that it is linear and continuous, or you give us the continuity also. It's only a minor mistake in the statement,but I understood the question – John D Aug 18 '17 at 11:20

1 Answers1

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Your argument is a little hard to follow. Try to write in complete sentences. State for which domains do the inequalities hold. For instance, in $(1)$ and $(2)$, do they work for every $n$ or some particular $n$? Also, you don't seem to use the hypothesis that the $(T_n)$ are uniformly bounded. If not, then how are you getting $(1)$ and $(2)$? Near inequality $(3)$, does your $n_0$ depend on $j$? What is the purpose op picking $k$ at the end? Here's a similar epsilonic argument.

Proof:

Fix $x\in B$ and let $\varepsilon>0$. Set $M:=\sup_n\|T_n\|<\infty$. Due to the density of $D$, there exists a sequence $(x_j)$ in $D$ such that $x_j\to x$. Then there is $j_0$ such that every $n\in\mathbb{N}$ and $j\geq j_0$ satisfies $$ \|T_nx_j - T_nx\| \leq \|T_n\|\|x_j-x\| \leq M\|x_j - x\| < \varepsilon/3. \tag{1}$$

Since the sequence $(x_j)$ consists of elements in $D$ and $T_nd\to Td$ for all $d\in D$, we have that for each $j$ there exists $n_j$ such that if $n\geq n_j$, then $$ \|T_nx_j-T_nx\| < \varepsilon/3. \tag{2}$$

Because $T$ is bounded (assuming this), then there is $j_1$ such that $j\geq j_1$ implies $$ \|Tx_j-Tx\| < \varepsilon/3. \tag{3} $$

If $j\geq\max\{j_0,j_1\}$ and $n\geq n_j$, then we obtain from $(1)$, $(2)$, and $(3)$ that

\begin{align*} \|T_nx-Tx\| &\leq \|T_nx-T_nx_j\|+\|T_nx_j-Tx_j\|+\|Tx_j-Tx\| \\ &< \varepsilon/3 + \varepsilon/3 + \varepsilon/3 \\ &= \varepsilon \tag*{$\square$} \end{align*}

Here's almost the same argument, except this one avoids all $\varepsilon$'s.

Proof:

Suppose $x\in B$. By the fact that $D$ is dense there exists a sequence $(x_k)$ in $D$ such that $x_k\to x$. Set $M:=\sup_{n}\|T_n\|$, which exists by hypothesis. Then, assuming that $T$ is also bounded and linear, we obtain \begin{align*} \|T_n x - Tx\| &\leq \|T_nx - T_nx_k\|+\|T_nx_k-Tx_k\|+\|Tx_k-Tx\| \\ &\leq M\|x-x_k\| + \|T_nx_k-Tx_k\|+\|T\|\|x_k-x\| \end{align*} for every $k\in\mathbb{N}$. Taking the the limit supremum of the above inequality as $n\to\infty$ yields $$ \limsup_{n\to\infty}\|T_nx-Tx\| \leq M\|x-x_k\|+\|T\||x_k-x|| $$ for every $k\in\mathbb{N}$, because $\|T_nx_k-Tx_k\|\to0$ as $n\to\infty$ by hypothesis. Finally, taking $k\to\infty$ implies $$ \limsup_{n\to\infty}\|T_nx-Tx\|\leq \lim_{k\to\infty}(M\|x-x_k\|+\|T\|\|x_k-x\|)=0. $$ Therefore $T_nx\to Tx$ for every $x\in B$. $\square$

In both proofs, there was no use of the assumption that $B$ is complete. We did however assume that $T$ was bounded and linear. Under certain hypothesis, the fact that $B$ is complete could imply that $T$ is bounded and linear on $B$. If you can clear up what we are allowed to assume about $T$, then I'll happily edit my post and add this part.

John Griffin
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  • @JohnGriffin..how do we know that $T$ is bounded? – Marios Gretsas Aug 17 '17 at 15:14
  • @MariosGretsas The question doesn't clearly state anything about $T$. I'm not sure if we can even assume that it is linear. From the hypothesis that $T_n\to T$ pointwise on $D$ and that $(T_n)$ are linear, we can deduce that $T(\alpha x+\beta y)=\alpha Tx+\beta Ty$ whenever $x,y,\alpha x+\beta y$ are all in $D$. But $D$ is only assumed to be a subset, so there is no guarantee that $x,y\in D$ even implies $\alpha x+\beta y\in D$. – John Griffin Aug 17 '17 at 15:23
  • We can prove that $T$ is continuous in D but is it enough for $T$ to be continuous in the whole space? – Marios Gretsas Aug 17 '17 at 15:25
  • @MariosGretsas Consider $B=\mathbb{R}$, $D=\mathbb{Q}$, $T_n=0$, and $T=\chi_{B\setminus D}$. Then $B$ is Banach, $D$ is dense in $B$, $(T_n)$ are linear and uniformly bounded, and $T_n\to T$ pointwise on $D$. But $T$ isn't linear nor bounded, and $T_n\not\to T$ pointwise on $B$. This shows that there are assumptions missing about $T$. – John Griffin Aug 17 '17 at 15:38
  • Indeed the O.P must be more clear – Marios Gretsas Aug 17 '17 at 15:44
  • @JohnGriffin Thanks for the tips. But what's wrong with the following reasoning: Let $T_n \rightarrow T$ pointwise on D and ($T_n$) be uniformly bounded by $C > 0$ so that $||T_nx|| \leq C||x||$ $\forall x \in D$ $\forall n \in \mathbb{N}$. Then, given $\epsilon >0$ we find $n_0$, so that $||Tx - T_nx|| \leq \epsilon \ \forall n>n_0$. Then $||Tx|| =||Tx - T_nx+T_nx|| \leq ||Tx - T_nx|| + ||T_nx|| \leq \epsilon + C||x||$ for all x, so T must be bounded on D. – JtSpKg Aug 18 '17 at 15:53
  • @JtSpKg Nothing is wrong there. However it's a little weird to say bounded on $D$ because (this version of) boundedness is only valid for linear maps, and we can't say that $T$ is linear on $D$ because $D$ may not be a subspace. Even so, I don't see how this would give that $T$ is bounded and linear on all of $B$. In fact, the example that is a couple of comments back shows that we need additional hypothesis. – John Griffin Aug 18 '17 at 19:46