Let $A$ be a Banach space with $\{f_n\}_{n=1}^\infty$ a sequence of operators in $B(A)$ such that $\lim\limits_{n\rightarrow\infty}\phi(f_n a)$ exists for all $\phi$ in the dual of $A$ and all $a\in A$.

How do I show that:

  1. $\sup_{n\geq 1}||f_n||<\infty$,
  2. the map $M:A^*\rightarrow A^*$ with $M\phi(a)=\lim\limits_{n\rightarrow\infty}\phi(f_n a)$ is in $B(A^*)$,
  3. $A$ is reflexive $\implies$ there exists a unique bounded linear operator $F$ such that $\phi(Fa)=\lim\limits_{n\rightarrow\infty}\phi(f_n a)$ for all $\phi\in A^*$ and all $a\in A$.

What I thought:
1. The uniform boundedness principle says that (for $X,Y$ Banach spaces, $S$ a non-empty set, $s\in S$, $f_s\in B(X,Y)$) if for all $x\in X$ the set $\{||f_s(x)||:s\in S\}$ is bounded, then $\{||f_s||:s\in S\}$ is bounded too.
How can I use this here? I was told to apply the principle twice.

2.The linearity seems clear $$M(\phi+\psi)(a)=\lim\limits_{n\rightarrow\infty}(\phi+\psi)(f_n a)=\lim\limits_{n\rightarrow\infty}(\phi(f_n a)+\psi(f_n a))=M\phi(a)+M\psi(a).$$ For the boundedness, I need to show that $||M\phi||_{A^*}\leq K ||\phi||_{A^*}$ for a constant $K\geq 0$, where $||\phi||_{A^*}=\sup\{||\phi(a)||_A:||a||_A\leq 1\}$ and $||M\phi||_{A^*}=\sup\{\lim\limits_{n\rightarrow\infty}||\phi(f_n a)||_A:||a||_A\leq 1\}$.
How do I show $||M\phi||_{A^*}\leq K ||\phi||_{A^*}$?

  1. Let $\alpha_a\in A^{**}$ be the evaluation map and $J:A\rightarrow A^{**}$ the map that sends every element of $A$ to its evaluation map. So $J(a)(f)=f(a)$ for $f\in A^*$.
    If $J(A)=A^{**}$, then $A$ is reflexive. How do I use this to prove 3.?
  • $\begingroup$ Ad 1, Consider $f_n(a)$ as an element of $A^{\ast\ast}$. $T_n(\phi) = \phi(f_n(a))$ for $\phi \in A^{\ast}$. Ad 2, It must be $\lvert \phi(f_na)\rvert$, since $\phi(f_na)\in \mathbb{C}$. Use properties of operator norms and part 1 to get a uniform (in $n$) bound for $\lvert \phi(f_na)\rvert$ in terms of $\lVert\phi\rVert_{A^{\ast}}$ and $\lVert a\rVert_A$. Ad 3, think of $M^{\ast}$. $\endgroup$ – Daniel Fischer Dec 7 '16 at 13:30
  • $\begingroup$ @DanielFischer 1. Then $f_n(a)$ sends every $\phi\in A^*$ to $T_n(\phi)$, I don't see the connection with the uniform boundedness principle. $\endgroup$ – user395531 Dec 7 '16 at 13:48
  • $\begingroup$ $X = A^{\ast}, Y = \mathbb{C}$, and $\{ \phi \mapsto \phi(f_na) : n \in \mathbb{N}\}$ as the family of linear operators. $\endgroup$ – Daniel Fischer Dec 7 '16 at 13:51
  • $\begingroup$ @DanielFischer So $\{||T_n(\phi)||\}$ is bounded since all $\phi$ are bounded, giving that $\{||T_n||\}$ is bounded? $\endgroup$ – user395531 Dec 7 '16 at 13:55
  • $\begingroup$ For every $\phi \in A^{\ast}$, $\{ T_n(\phi) : n \in \mathbb{N}\}$ is bounded. By Banach-Steinhaus, $\{ \lVert T_n\rVert_{A^{\ast\ast}} : n \in \mathbb{N}\}$ is bounded. But $J \colon A \to A^{\ast\ast}$ is an isometry, and $T_n = J(f_na)$, so $\{ \lVert f_n a\rVert_A : n \in \mathbb{N}\}$ is bounded. $\endgroup$ – Daniel Fischer Dec 7 '16 at 14:01

To obtain $\sup\limits_{n\geqslant 1} \lVert f_n\rVert < +\infty$ from the uniform boundedness principle, we need that for every $a \in A$ the set $\{ f_n(a) : n \geqslant 1\}$ is bounded, i.e. $\sup\limits_{n\geqslant 1} \lVert f_n(a)\rVert < +\infty$. We obtain that from the premises that for every $a\in A$ and $\phi\in A^{\ast}$ the sequence $\phi(f_n(a))$ converges (in $\mathbb{C}$ or $\mathbb{R}$, whatever the scalar field is). For, if we fix $a\in A$ and define $T_n = J(f_n(a)) \in A^{\ast\ast}$, the premise is that for every $\phi \in A^{\ast}$ the sequence $T_n(\phi)$ is convergent, and therefore bounded. Since $A^{\ast}$ is a Banach space, the uniform boundedness principle asserts

$$\sup_{n \geqslant 1} \lVert T_n\rVert < +\infty.\tag{1}$$

But $J \colon A \to A^{\ast\ast}$ is an isometry, and thus $(1)$ gives us

$$\sup_{n \geqslant 1} \lVert f_n(a)\rVert < +\infty,\tag{2}$$

and since $A$ is a Banach space, this is just what we need to obtain the desired

$$K := \sup_{n \geqslant 1} \lVert f_n\rVert < +\infty\tag{3}$$

from the uniform boundedness principle.

The premise that for every $a\in A$ and $\phi \in A^{\ast}$ the sequence $\phi(f_n(a))$ converges can be stated as

For every $\phi \in A^{\ast}$ sequence $(\phi \circ f_n)$ converges pointwise on $A$.

Let's call the limit function $M(\phi)$. First we note that as a pointwise limit of linear maps, $M(\phi)$ is linear too -

\begin{align} M(\phi)(a+b) &= \lim \phi(f_n(a+b)) = \lim \phi(f_n(a) + f_n(b))\\ &= \lim \bigl(\phi(f_n(a)) + \phi(f_n(b))\bigr) = \lim \phi(f_n(a)) + \lim \phi(f_n(b)) \end{align}

and so on. Next we see that $M(\phi)$ is continuous, so $M(\phi) \in A^{\ast}$, for every $\phi \in A^{\ast}$, and find a bound for its norm. Fix an arbitrary $\phi \in A^{\ast}$. For every $n \geqslant 1$ and every $a \in A$, we have

$$\lvert \phi(f_n(a))\rvert \leqslant \lVert \phi\rVert\cdot \lVert f_n\rVert \cdot \lVert a\rVert \leqslant K\lVert\phi\rVert\cdot\lVert a\rVert\tag{4}$$

from the definition of the norms on $A^{\ast}$ and $B(A)$ and formula $(3)$. It follows that

$$\lvert (M\phi)(a)\rvert = \left\lvert\lim_{n\to\infty} \phi(f_n(a)) \right\rvert = \lim_{n\to\infty} \lvert \phi(f_n(a))\rvert \leqslant K\lVert\phi\rVert\cdot\lVert a\rVert,$$

and that implies that $M(\phi)$ is indeed continuous, and

$$\lVert M(\phi)\rVert = \sup \{\lvert M(\phi)(a)\rvert : \lVert a\rVert \leqslant 1\} \leqslant \sup \{K\lVert\phi\rVert\cdot\lVert a\rVert : \lVert a\rVert \leqslant 1\} \leqslant K \lVert\phi\rVert.\tag{5}$$

After checking that the map $M \colon \phi \mapsto M(\phi)$ is linear, $(5)$ immediately shows $M \in B(A^{\ast})$, and $\lVert M\rVert \leqslant K$.

For point 3, consider the Banach-space-adjoint, dual, or transpose of $M$, that is the map $M^{\ast} \colon A^{\ast\ast} \to A^{\ast\ast}$ given by $M^{\ast}(\xi) = \xi \circ M$ for $\xi \in A^{\ast\ast}$. If $A$ is reflexive, we can define $F \in B(A)$ by

$$F = J^{-1} \circ M^{\ast} \circ J.\tag{6}$$

Then for $a\in A$ and $\phi \in A^{\ast}$ we have

$$\phi(Fa) = \phi(J^{-1}(M^{\ast}(Ja)) = (M^{\ast}(Ja))(\phi) = (Ja)(M\phi) = (M\phi)(a) = \lim \phi(f_na),\tag{7}$$

which is the desired property. It remains to show the uniqueness of $F$. We show that generally for $T,S \in L(X,Y)$, where $X,Y$ are normed spaces, we have

$$T^{\ast} = S^{\ast} \iff T = S,\tag{8}$$

where $T^{\ast}\in L(Y^{\ast}, X^{\ast})$ is the transpose (dual, adjoint) of $T$ given by $T^{\ast}(\lambda) = \lambda \circ T$ (and analogously for $S$). By the linearity of $T\mapsto T^{\ast}$ (that is, $(aS + bT)^{\ast} = a(S^{\ast}) + b(T^{\ast})$ for $S,T \in L(X,Y)$ and $a,b\in \mathbb{C}$), $(8)$ is equivalent to

$$T^{\ast} = 0 \iff T = 0.\tag{9}$$

So suppose that we have $T \in L(X,Y)$ with $T^{\ast} = 0$. Then for every $x\in X$ and $\lambda \in Y^{\ast}$ we have

$$0 = (T^{\ast}\lambda)(x) = \lambda(Tx).$$

By Hahn-Banach, this implies $Tx = 0$ - otherwise there'd be a $\lambda \in Y^{\ast}$ with $\lambda(Tx) \neq 0$ - and $Tx = 0$ for all $x\in X$ just means $T = 0$.

  • $\begingroup$ Why does $T^*=S^*\iff T=S$ give the uniqueness of $F$? $\endgroup$ – user395531 Dec 10 '16 at 20:28
  • 1
    $\begingroup$ The condition that for all $a$ and $\phi$ we should have $\phi(Fa) = \lim \phi(f_n a)$ translates to $F^{\ast} = M$. So if we have two such operators $F_1,F_2$, that means $F_1^{\ast} = F_2^{\ast}$. With $T^{\ast} = S^{\ast} \iff T = S$, we deduce $F_1 = F_2$. So the condition characterises $F$ uniquely. $\endgroup$ – Daniel Fischer Dec 10 '16 at 20:32
  • $\begingroup$ And why does linearity give us that we want $T^*=0\iff T=0$? Because $\{T^*=S^*\iff T=S\}\iff\{T^*-S^*=0\iff T-S=0\}\iff\{(T-S)^*=0\iff T-S=0\}$? $\endgroup$ – user395531 Dec 10 '16 at 20:38
  • $\begingroup$ Yes, exactly that. $\endgroup$ – Daniel Fischer Dec 10 '16 at 20:44

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