This is from Tao's book;

Let $X$ be Banach space, let Y be normed vector space, and let $(T_n)_{n=1}^{\infty}$ be a family of continuous linear operators $T_n : X \to Y$. Then following is equivalent.

(i) (Pointwise convergence) For every $x \in X$. $T_n(x)$ converges strongly in $Y$ as $n \to \infty$.

(ii) (Pointwise convergence to a continuous limit) There exists a continuous linear map $T:X \to Y$ such that for every $x\in X$, $T_n(x)$ converges strongly in $Y$ to $T(x)$ as $n\to\infty$.

(iii) (Uniform boundedness + dense subclass convergence) The operator norms $\{||T_n|| : n = 1, 2, . . .\}$ are bounded, and for a dense set of $x$ in $X$, $T_n(x)$ converges strongly in $Y$ as $n \to \infty$

What I have trouble is $(i) \to (iii)$ and $(iii) \to (ii)$.

My proof on $(i) \to (iii)$ is like below; second statement of $(iii)$ is easy so it suffices to prove only boundedness of the set of operator norms. Since convergent sequence is bounded, $\{ T_n(x)\}_{n=1}^{\infty}$ is clearly bounded sequence. Hence there is some $N $ such that $||T_n(x)|| < N$. Since $T_n$ is linear, $||T_n(x)|| = ||T_n ||_{\mathrm{op}}||x|| < N$, and since $x$ is fixed, $||T_n||_{\mathrm{op}}$ is bounded. Is this argument reasonable? In the Tao's note, he mention Baire category theorem to prove this.But I don't know how to prove this using Baire category theorem directly.

For second question, $(iii) \to (ii)$, I have an idea that construct $T:X \to Y$ such that $T(x) =\lim_{n \to \infty} T_n(x)$ and $T(0)=0$. But I know that $T$ is linear and continuous on dense set, but don't know how to extend it to limit point of $X$. Could you give me some hint?


First of all, in order to have $(iii)$ be equivalent to $(i)$ and $(ii)$, we need that $Y$ is complete (a Banach space) too.

As a counterexample when $Y$ is not complete, consider $X = \ell^p$, and $Y$ the subspace of sequences with only finitely many nonzero terms. Then let

$$(T_nx)_k = \begin{cases}x_k &, k \leqslant n\\ 0 &, k > n. \end{cases}$$

$T_n$ is a bounded sequence of linear operators $X\to Y$, and $T_n(x)$ converges for all $x$ in the dense subspace $Y$ of $X$. But $T_n(x)$ does converge only for $x\in Y$, so neither $(i)$ nor $(ii)$ hold in this example.

Your attempt to prove $(i) \Rightarrow (iii)$ does not work, there are two problems. First, from $\lVert T_n(x)\rVert < N$, you cannot deduce $\lVert T_n\rVert_{op}\lVert x\rVert < N$, since you only have the inequality $\lVert T_n(x)\rVert \leqslant \lVert T_n\rVert_{op}\lVert x\rVert$, not equality.

Second, the bound on the sequence $T_n(x)$ depends on $x$, you only have $\lVert T_n(x)\rVert \leqslant N(x)$.

To show the uniform boundedness, you consider the closed sets $$A_K = \{x \in X : \lVert T_n(x)\rVert \leqslant K \text{ for all } n\}.$$

Since $T_n(x)$ is bounded for every $x\in X$, you have

$$X = \bigcup_{k=1}^\infty A_k.$$

From Baire's theorem, deduce that some $A_k$ is a neighbourhood of $0$ in $X$.

For the implication $(iii) \Rightarrow (ii)$ (under the assumption that $Y$ is complete!), note that

  • the set $C = \left\{ x \in X : \lim\limits_{n\to\infty} T_n(x)\text{ exists}\right\}$ is a linear subspace of $X$,
  • the map $T\colon C \to Y$ defined by $T(x) = \lim\limits_{n\to\infty} T_n(x)$ is linear and continuous.

Then the general theory asserts the existence of a continuous (linear) extension $\tilde{T}\colon \overline{C}\to Y$. Since $C$ is by assumption dense, $\overline{C} = X$. Now use the boundedness of the norms - the equicontinuity of the family $(T_n)$ - to deduce that actually $C = X$.


For (i) $\implies$ (iii). For each $N=1,2,3,\ldots,$ define $F_{N}=\{ x : \sup_{n}\|T_{n}x\| \le N \}$, and show every $x$ is in some $F_{N}$.

For (iii) $\implies$ (ii). If $\|T_{n}\|\le M$ for some $M$ and all $n$, $$ \begin{align} \|T_{n}x-T_{n+k}x\| & \le \|T_{n}x-T_{n}y\|+\|T_{n}y-T_{n+k}y\|+\|T_{n+k}y-T_{n+k}x\| \\ & \le 2M\|x-y\|+\|T_{n}y-T_{n+k}y\|. \end{align} $$

  • $\begingroup$ Thank you for simple proof! $\endgroup$
    – user124697
    Feb 26 '14 at 7:21

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