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I would really appreciate it if you could give me some advice on the part (d) 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}

(a) 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.

(b) Prove that $id: (C, \tau) \rightarrow (C, \sigma)$ is nevertheless not continuous, although it is sequentially continuous (by Lebesgue's dominated convergence theorem). Hence $(C, \tau)$ is not metrizable. Show also directly that $(C, \tau)$ has no countable local base.

(c) Prove that every continuous linear functional on $(C, \tau)$ is of the form \begin{equation} f \rightarrow \sum_{i=1}^n c_i f(x_i) \end{equation} for some choice of $x_1, \ldots, x_n$ in $[0, 1]$ and some $c_i \in \mathbb{C}$.

(d) Prove that $(C, \sigma)$ contains no convex open sets other than $\varnothing$ and $C$.

First part of this question is here, the second part is here and the third part is here. I have no idea solving this part too.

Thanks in advance.

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Hossein Moradi
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1 Answers1

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The same thing holds for the spaces $L^p$, $0 < p < 1$, and Rudin gives an elegant proof for this in his book (section 1.47). Pretty much the same argument works for $C[0,1]$ with the given metric (some boring details need to be added).

Assume that $\emptyset \neq V \subset C$ is open and convex and $0 \in V$. From this it follows that the open ball $B(0,r)$ is contained in $V$ for some $r > 0$. Let $f \in C$ be arbitrary. Because $d(fn,0) \leq 1$ we get $n^{-1} d(fn,0) \to 0$ as $n \to \infty$. Thus there exists $n \in \Bbb{N}$ such that $$ n^{-1} d(fn,0) < r\,. $$ Now, because the map $$ h \mapsto \int_0^x \frac{|h(t)|}{1 + |h(t)|} \, dt $$ is continuous for each $h \in C$, there exists points $\{x_0, ..., x_n\} \subset [0,1]$ such that $$ \int_{x_{i-1}}^{x_i} \frac{|nf(x)|}{1+|nf(x)|} = n^{-1} \int_0^1 \frac{|nf(x)|}{1+|nf(x)|} = n^{-1} d(nf,0)\,. $$ Make sure you understand why the points $x_i$ exist!

Now define functions $g_i$ by $g_i(x) = nf_i(x)$ on $[x_{i-1}, x_i]$ and $0$ otherwise, $1 \leq i \leq n$. Then we have $$ \int_0^1 \frac{|g_i(x)|}{1 + |g_i(x)|} = n^{-1} d(fn,0) < r\,. $$

Now it would be nice to conclude that $g_i \in B(0,r)$ and $f = \sum n^{-1} g_i$, but clearly we don't know if $g_i \in C$. But clearly we can modify the functions $g_i$ in a small enough set such that they become continuous and the integral $\int_0^1 \frac{|g_i(x)|}{1 + |g_i(x)|}$ is still less than $r$. Then the modified $g_i$ belong to $B(0,r) \subset V$, and because $V$ is convex the convex combination $g := \sum_{i=1}^n n^{-1} g_i$ belongs to $V$.

Now I argue that the modifying of $g_i$ can be done in such a way that $g = f$. For example, take the point $x_1$. This is where the functions $g_1$ and $g_2$ "meet". For clarity I assume that $x_1 = 0$ even though in our situation it never is $0$. Then redefine the functions on $[- \varepsilon , \varepsilon]$ as

\begin{align} g^*_1(x) &= \begin{cases}g_1(x) , & x \in [x_0, -\varepsilon] \\ \frac{1 - x/\varepsilon}{2}g_1(x), & x \in (-\varepsilon, 0], \\ \frac{1 - x/\varepsilon}{2} g_2(x)\, & x \in (0, \varepsilon] \end{cases} \\ g^*_2(x) &= \begin{cases}g_2(x) , & x \in [\varepsilon, x_2] \\ \frac{1 + x/\varepsilon}{2}g_2(x), & x \in [0, \varepsilon),\\ \frac{1 + x/\varepsilon}{2} g_1(x)\, & x \in [- \varepsilon, 0) \end{cases} \end{align}

Now we see that $g_1^* + g_2^* = g_1 + g_2 = nf$ on $[- \varepsilon, \varepsilon]$. Hence $g = f$ and the claim holds if $0 \in V$ since $f$ was arbitrary. I think the case $0 \notin V$ follows from the metric $d$ being translation invariant.

So as you see, I had to add some nasty details. I recommend you to read the proof for $L^p$, $0 < p < 1$, from Rudin's book since it's very simple and thus it's easier to understand the main arguments. Also, I think there exists a simpler proof but I haven't come up with one yet.

desos
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