Suppose that $\{T_j\}_{1}^{\infty}$ is a sequence of bounded linear transformations from a normed vector space $X$ into a Banach space $Y$, suppose $\lVert T_j\rVert\leq M < \infty$ for all $j$ and suppose there is a dense set $E\subset X$ such that $\{T_k(x)\}_{1}^{\infty}$ converges for every $x\in E$. Prove that $\{T_j(x)\}_{1}^{\infty}$ converges for $x\in X$.

Attempted proof: Let $x\in X$. We know $Y$ is complete, therefore if we show $\{T_j(x)\}_{1}^{\infty}$ is Cauchy then we are done.

Let $\epsilon > 0$, choose an $N$ such that $$\lVert T_m(x) - T_n(x) \rVert < \epsilon \ \forall m,n\geq N$$ Since $E$ is dense in $X$ and $\{T_j(x')\}$ is convergent, we can choose an $x'\in E$ with $\lVert x - x'\rVert < \frac{\epsilon}{3M}$, and find an $N$ such that $$\lVert T_m(x') - T_n(x')\rVert < \frac{\epsilon}{3} \ \forall m,n\geq N$$ Then \begin{align*} \lVert T_m(x) - T_n(x)\rVert &\leq \lVert T_m(x - x')\rVert + \lVert T_m(x') - T_n(x')\rVert + \lVert T_n(x' - x)\rVert\\ &< 2M\lVert x - x'\rVert + \frac{\epsilon}{3}\\ &< \frac{2\epsilon}{3} + \frac{\epsilon}{3}\\ &= \epsilon \end{align*}


Let $x\in X$. Then there is $y\in E$ so that $\|x-y\|_X$ is small. Note

$$ \begin{split} \| T_j x - T_k x\|_Y &= \| (T_j -T_k)x \| \\ &\le \|(T_j - T_k) (x-y) \|_Y + \|T_j y- T_ky \|_Y \\ &\le \| T_j - T_k\| \cdot \|x-y\|_X + \|T_j y- T_ky \|_Y \end{split} $$

Can you finish from here?

  • $\begingroup$ I don't really follow yet $\endgroup$
    – Wolfy
    Feb 25 '16 at 4:31
  • $\begingroup$ Like you said, you want to show that $\{T_j x\}$ is Cauchy. Now do you know how to show that the right hand side will be arbitrarily small? @MorganWeiss $\endgroup$
    – user99914
    Feb 25 '16 at 4:33
  • $\begingroup$ no, I do not know $\endgroup$
    – Wolfy
    Feb 25 '16 at 4:41
  • $\begingroup$ Note that $\{T_j y\}$ is Cauchy. So can you deal with the term $\|T_j y - T_k y\|_Y$? @MorganWeiss $\endgroup$
    – user99914
    Feb 25 '16 at 4:51
  • $\begingroup$ yea, I appreciate your help but I am not use to the type of notation your are using, so I cannot follow your reasoning unless you do a similar approach with different notations $\endgroup$
    – Wolfy
    Feb 25 '16 at 19:35

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