Part (a) of Exercise 13 of first chapter of Rudin's book "Functional Analysis"

Question

I would really appreciate it if you could give me some advice on the part (a) of Exercise 13 of first chapter of Walter Rudin's book "Functional Analysis":

Let $C$ be the vector space of all complex continuous functions on $[0, 1]$. Define \begin{equation} d(f,g) = \int_0^1 \frac{\lvert f(x) - g(x) \rvert}{1 + \lvert f(x) - g(x) \rvert} \ dx \ . \end{equation} Let $(C, \sigma)$ be $C$ with the topology induced by this metric. Let $(C, \tau)$ be the topological vector space defined by the semi-norms \begin{equation} P_x(f) = \lvert f(x) \rvert, \qquad (0 \leq x \leq 1), \end{equation} Prove that every $\tau$-bounded set in $C$ is also $\sigma$-bounded and that the identity map $id: (C, \tau) \rightarrow (C, \sigma)$ therefore carries bounded sets into bounded sets.

I tried using the theorem that says a set $E \subseteq C$ is bounded if and only if every semi-norm in our semi-norms is bounded on $E$. This theorem tells us that if $E$ be a bounded set in $(C, \tau)$, then for every $x \in [0, 1]$, $P_x(E)$ is bounded, i.e. \begin{equation} \forall x \in [0, 1] \ \exists M_x, \quad s.t. \quad \forall f \in E, \quad \lvert f(x) \rvert \leq M_x \ . \end{equation} I think now we should use Uniform boundedness principle and obtain $M > 0$ such that \begin{equation} \forall x \in [0, 1] \ \forall f \in E, \quad \lvert f(x) \rvert \leq M \ . \end{equation} Then we have $d(f, 0) \leq \frac{M}{1+M}$ for all $f \in E$. So $E$ is bounded in $(C, \sigma)$.

In the last step, to use Uniform boundedness principle, I think we should prove that $(C, \tau)$ is a Banach space, and $(C, \sigma)$ a normed space. I don't know what should I do in this step.

Answer 1

What's condition of UBP and what's the conclusion? You need a Banach space, a normed linear space and a linear map between them. Your conclusion is also about the norm of a bounded linear map. Where is the linear map in your question?

Even if you define those $Q_x$, they are linear functionals in the dual space of $C[0,1]$ with the $\|\cdot\|_\infty$ norm. Nothing about these $f\in E$ can you obtain from the boundedness of these functionals.

Besides, what's the definition of a bounded set in some topological vector space? It is not equivalent to a bounded set with respect to some metric. Actually the whole space is already bounded with respect to the metric.

If $B_r(0)$ is an open $0$-nbhd in ($C,\sigma)$, you want to show $\exists t>0$ s.t. $E\subseteq t\cdot B_r(0)$. For some $f\in E$, $f=t\cdot\frac{f}{t}.$ Your goal is to prove $\frac{f}{t}\in B_r(0)$. i.e. $$ \int_0^1 \frac {\frac{|f(x)|}{t}} {1+\frac{|f(x)|}{t}} dx = \int_0^1 \frac{|f(x)|}{t+|f(x)|}dx<r. $$ That is to find some $t$ big enough such that $\int_0^1\frac{|f(x)|}{t+|f(x)|}dx$ can be arbitrarily small for pointwise bounded family of functions. When $t\to\infty$, $\frac{|f(x)|}{t+|f(x)|}\to0$ pointwise and $\frac{|f(x)|}{t+|f(x)|}$ is always bounded by an integrable function. What does this imply? Can you prove that $$ \lim_{t\to\infty}\int_0^1 \frac{|f(x)|}{t+|f(x)|}dx=0? $$ How to make that happen for all $f\in E$ so that you can have the $t$ big enough uniformly? i.e. the following hidden hint.

The pointwise supreme of any family of lower semi-continuous functions is also lower semi-continuous thus measurable.

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Define $\phi(x)=\sup_{f\in E}|f(x)|$. They are all bounded thus $\phi$ is a well defined real function and $\phi$ is lower semi-continuous. Thus $$d(f/t,0)=\int_0^1\frac{|f(x)|}{t+|f(x)|}dx\le\int_0^1\frac{\phi(x)}{t+\phi(x)}dx\to 0\ \ (DCT).$$

Answer 2

You want to apply the UBP to the functionals given by the point evaluations, say $Q_x (f)=f (x) $. So your $Y $ ( in the notation from the linked article in Wikipedia ) is $\mathbb C $.

And the topology on $C $ plays no role in your argument: you are already assuming that, for each $f\in E $, $|Q_x (f)|\leq M_x$. So you apply the UBP to the functionals $\{Q_x\}_{x\in [0,1]} $ acting on the Banach space $C [0,1] $.