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These days I am doing some independent study of functional analysis. While solving problems, I could not handle the following part of an exercise (exercise 13, chapter 1 of Rudin's Functional Analysis).

Let $C$ be the vector space of all complex continuous functions on $[0,1]$. Define $$d(f,g)=\int_0^1 \frac{|f(x)-g(x)|}{1+|f(x)-g(x)|}dx.$$

Let $(C,\sigma)$ be $C$ with the topology induced by this metric. Let $(C,\tau)$ be the topological vector space defined by the seminorms $$p_x(f)=|f(x)| \quad (0\le x \le 1).$$

Prove that every $\tau-$bounded set in $C$ is also $\sigma-$bounded and that the identity map $id:(C,\tau)\to (C,\sigma)$ therefore carries bounded sets into bounded sets.

*Because of the definition of our seminorm, a set $E$ is $\tau-$bounded if and only if for each $x\in [0,1]$ there exists $M_x \ge 0$ such that $p_x(f)=|f(x)| \le M_x$ for all $f\in E$.

To show that $E$ is $\sigma-$bounded, I tried to prove that for any $\delta>0$, $B(0,\delta)$ absorbs $E$, i.e. there exists $t>0$ such that $\frac{1}{t} E \subset B(0,\delta)$, but I am stuck here.

I would really appreciate it if you could give me some sketches or hints on this. Thank you.

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1 Answer 1

up vote 1 down vote accepted

Sketch: Define $b_E(x) := \sup\limits_{f \in E} \lvert f(x)\rvert$. $b_E$ is the supremum of continuous functions, hence it is lower semicontinuous, hence measurable.

Apply the dominated convergence theorem to $\frac1t b_E$.

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Thanks a lot. I mistakenly thought that the sup function might not be measurable! Thanks again. –  user.so Jul 20 '13 at 18:27

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