Problem 6

Exercise 5.6 (Sequence of Bounded Operators on a Banach space)  

1. Let $\{A_n:X\to X\}_{n=1}^\infty$ be a sequence of bounded linear operators on a Banach space $X$ such that $A_n x$ converges for every $x\in X$. Show the following operator on $X$ is bounded:

   $$Ax := \lim_{n\to\infty} A_n x.$$ (5.25)

2. Can the same conclusion be drawn if $X$ is not a Banach space?

Proof. We can argue that $\Vert A\Vert<\infty$ using uniform boundedness. The convergence hypothesis implies that

$$\sup_{n\geq 1} \Vert A_nx\Vert <\infty \quad \forall x \in X$$ (5.26)

Uniform boundedness says

$$\sup_{n\geq 1}\Vert A_n\Vert<\infty$$ (5.27)

We will use this estimate to prove that

$$\Vert A\Vert = \sup_{\Vert x\Vert=1}\lim_{n\to\infty}\Vert A_n x\Vert < \infty$$ (5.28)

The limit within the supremum always exists by the convergence hypothesis, so for any $x$ it is true that

$$\lim_{n\to\infty}\Vert A_n x\Vert = \liminf_{n\to\infty}\Vert A_n x\Vert = \sup_{n\geq 1}\inf_{m\geq n} \Vert A_m x\Vert$$ (5.29)

This is substituted into the equation for $\Vert A\Vert$, and we interchange some limits to see

$$\Vert A\Vert = \sup_{\Vert x\Vert=1}\lim_{n\to\infty}\Vert A_n x\Vert = \sup_{\Vert x\Vert=1}\sup_{n\geq 1}\inf_{m\geq n} \Vert A_m x\Vert$$ (5.30)

$$= \sup_{n\geq 1}\sup_{\Vert x\Vert=1}\inf_{m\geq n} \Vert A_m x\Vert$$ (5.31)

$$\leq \sup_{n\geq 1}\inf_{m\geq n}\sup_{\Vert x\Vert=1} \Vert A_m x\Vert$$ (5.32)

$$= \sup_{n\geq 1}\inf_{m\geq n} \Vert A_m\Vert$$ (5.33)

$$\leq \sup_{n\geq 1} \Vert A_n\Vert$$ (5.34)

$$< \infty$$ (5.35)

Consider the sequence of bounded operators

$$T_n : C(\mathbb{R}) \to \mathbb{R}$$ (5.36)

$$f \mapsto \int_{-n}^n f(x)dx$$ (5.37)

Each integration $T_n$ is over a compact domain, so the operators are bounded. But the limit operator is integration over the whole real line, which is unbounded, for example in the case of constant functions. $\qedsymbol$