3
$\begingroup$

I'm self-studying functional analysis. The following is from Rudin's "Functional Analysis, 2nd edition". It consists of parts from question 7 and 13 from the first chapter. I am not sure if my answers are correct, and would appreciate any hints/help. Thanks!

7) Let $X$ be the vector space of all complex functions on the unit interval $[0,1]$, topologized by the family of semi-norms

\begin{equation} p_x(f) = |f(x)|\,\,\,\,\,\, 0\leq x\leq 1 \end{equation}

Show that there is a sequence $\{f_n\}$ in $X$ such that $\{f_n\}$ converges to $0$ as $n\rightarrow \infty$, but if $\{\gamma_n\}$ is any sequence of scalars such that $\gamma_n \rightarrow \infty$, then $\{\gamma_nf_n\}$ does not converge to $0$ (Hint: use the fact that the collection of all complex sequences converging to $0$ has the same cardinality as $[0,1]$.)

13) Let $C$ be the vector space of all complex continuous functions on $[0,1]$. Define

\begin{equation} d(f,g) = \int_0^1\frac{|f(x) - g(x)|}{1+|f(x) - g(x)|}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) = |f(x)|\,\,\,\,\,0\leq x\leq1 \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 not continuous, although it is sequentially continuous. Hence $(C,\tau)$ is not metrizable. Show also directly that $(C,\tau)$ has no countable local base.

Attempt:

7) Im really struggling with this one. I don't see where the cardinality of $[0,1]$ comes in.

13) a) Let $E$ be a $\tau$-bounded set in $C$. Using Theorem 1.37 from the book, we can say that in the $\tau$-topology, a set $E\subset C$ is bounded if and only if every $p_x(f)$ is bounded on $E$. i.e., for any $f\in E$, $|f(x)| < k_x$ for all $x\in [0,1]$, where $k_x$ is some positive real number dependent on $x$.

If $g\in E$ then $d(g,0)<1$, because $|g(x)|<\infty$, from the above statement. Let $V_{\sigma} = \{g\in C : d(g,0) < 1\}$. Then, $tV_{\sigma} = \{f\in C : d(f,0)<t\}$. Therefore, if $h\in E$, then $h\in tV_{\sigma}$ for all $t\geq s$, for $s$ large enough, and so $E$ is also $\sigma$-bounded.

b) If $\{f_n\}\rightarrow f$ in the $\tau$-topology, then $d(f_n,f)\rightarrow 0$, and so $id$ is sequentially continuous.

To show $(C,\tau)$ does not have a countable base: Let $\mathcal{B}$ be the base constructed by the family of semi-norms. i.e., $\mathcal{B}$ is the collection of all finite intersections of sets $V(p_x,n) = \{f:p_x(f)<\frac{1}{n}\}$, $n$ a positive integer. For a real number $x$ and an integer $n$, a base element is given by:

\begin{equation} B_{x,n} = \{f\in C:|f(x)|<\frac{1}{n}\} \end{equation}

Because $[0,1]$ is uncountable, there is an uncountable number of base elements. We can therefore conclude that $id$ is not continuous (sequential continuity result, found in A6 in appendix of the book), and that $(C,\tau)$ is not metrizable.

$\endgroup$
0
9
$\begingroup$

Hint for exercise 7: Let $\mathcal{Z}$ be the set of complex sequences converging to $0$. Let $b\colon \mathcal{Z} \to [0,1]$ be a bijection. What might be a good choice for the value of $f_n(b(\zeta))$ then?

We choose $f_n(b(\zeta)) = \zeta_n$. If $(\gamma_n)$ is a sequence with $\gamma_n \to \infty$, define $\hat\zeta$ by $\hat\zeta_n = 1/\gamma_n$ if $\gamma_n\neq 0$, and $\hat\zeta_n = 0$ if $\gamma_n = 0$. Then $\gamma_n\cdot f_n(b(\hat\zeta)) = 1$ for all large enough $n$.

Concerning 13 a): No, the distance $d$ is not homogeneous,

$$t\cdot V_\sigma \neq \left\{ f\in C : d(f,0) < t\right\}.$$

If you define $\varphi(x) = \sup \{ \lvert f(x)\rvert : f \in B\}$ for a $\tau$-bounded subset $B\subset C$, what regularity properties of $\varphi$ do you know? And what has Lebesgue to say about the sequence $$\frac{\varphi/n}{1+\varphi/n}\,?$$

$\varphi$ is the supremum of continuous functions, hence it is lower semicontinuous, and therefore measurable. Also, $\varphi$ is non-negative and finite everywhere, since $B$ is $\tau$-bounded, so the sequence $\psi_n = \frac{\varphi/n}{1+\varphi/n} = \frac{\varphi}{n+\varphi}$ of measurable functions converges to $0$ pointwise, and it is evidently dominated by the constant function $1$. By Lebesgue's dominated convergence theorem, or by the monotone convergence theorem, it follows that $\int_0^1 \psi_n(x)\,dx \to 0$. Hence for every $r > 0$, there is an $n_r\in \mathbb{N}\setminus \{0\}$ such that $0 \leqslant \int_0^1 \psi_n(x)\,dx < r$ for all $n \geqslant n_r$. But that implies $n^{-1}\cdot B \subset B_r(0)$ for all $n\geqslant n_r$, so $B$ is shown to be $\sigma$-bounded.

Concerning 13 b): What is your argument for $d(f_n,f) \to 0$ when $f_n \to f$ pointwise?

The dominated convergence theorem. $0 \leqslant \delta_n(x) := \frac{\lvert f_n(x) - f(x)\rvert}{1+\lvert f_n(x) - f(x)\rvert} \leqslant 1$ for all $n$ and $x$, and $\delta_n \to 0$ pointwise. Thus the identity is sequentially continuous (which follows from the fact that it maps $\tau$-bounded sets to $\sigma$-bounded sets).

Why is $\operatorname{id} \colon (C,\tau) \to (C,\sigma)$ not continuous, although it is sequentially continuous?

Because not all $\sigma$-neighbourhoods of $0$ are $\tau$-neighbourhoods of $0$, see next point.

Can you find a neighbourhood $V$ of $0$ in $(C,\sigma)$ such that no $\tau$-neighbourhood of $0$ is contained in $V$?

Any $d$-ball $B_r(0)$ with $0 < r < 1$ will do. Since every $\tau$-neighbourhood of $0$ only constrains the values at finitely many points, every $\tau$-neighbourhood of $0$ contains functions with $d(f,0)$ arbitrarily close to $1$. Take a function that vanishes at the finitely many points controlled by the given $\tau$-neighbourhood, and has sufficiently high triangular spikes in between.

To show that $\tau$ has no countable local basis at $0$, it is not sufficient to expose one local basis that is uncountable. You must show that no countable system can be a neighbourhood basis of $0$.

How does a neighbourhood of $0$ in $\tau$ look, and hence, what can you say about a countable family of neighbourhoods? Which points of $[0,1]$ are controlled by any of these neighbourhoods?

A $\tau$-neighbourhood of $0$ is a set containing $T(\varepsilon; x_1,\dotsc,x_k) = \{ f \in C : \lvert f(x_i)\rvert < \varepsilon \text{ for } 1 \leqslant i \leqslant k\}$ for some $\varepsilon > 0$, non-negative integer $k$ and points $x_1,\dotsc,x_k \in [0,1]$. So a countable family $\mathscr{U} = \{U_n : n \in \mathbb{N}\}$ of neighbourhoods ($U_n \supset T(\varepsilon_n;x_1^n,\dotsc x_{k_n}^n)$) only constrains the values at countably many points. For any $x\in [0,1]\setminus \bigcup_n \{x_i^n : 1 \leqslant i \leqslant k_n\}$, none of the $U_n$ is contained in $T(1; x)$, so $\mathscr{U}$ is not a local base. Since $\mathscr{U}$ was an arbitrary countable family of neighbourhoods, it follows that any local base must be uncountable.

$\endgroup$
4
  • $\begingroup$ Regarding the statement that every $\tau$-neighbourhood of $0$ only constrains the values at finitely many points. Why is that if open sets are closed under arbitrary union? Couldn't you take a union of open sets as $x$ ranges over all of $[0,1]$? $\endgroup$
    – JessicaK
    Feb 5 '16 at 10:13
  • $\begingroup$ @JessicaK If you enlarge a set, you reduce the constraints imposed on the values of the functions belonging to the set. To constrain the values at more points, we intersect neighbourhoods. And only intersections of finitely many neighbourhoods are guaranteed to be neighbourhoods again. So with a countable family $(U_n)$ of neighbourhoods, we have only countably many points where an intersection of finitely many of the $U_n$ can impose a restriction on the values of the functions belonging to that intersection. All of these intersections only restrict the values at finitely many points, $\endgroup$ Feb 5 '16 at 10:32
  • $\begingroup$ but that finite set is a subset of the countable set of points where any of the $U_n$ imposes a restriction, that's why we have points where we can a priori know that no intersection of finitely many $U_n$ restricts the values of the functions belonging to that intersection. $\endgroup$ Feb 5 '16 at 10:34
  • $\begingroup$ I understand now. I completely misunderstood how enlarging a set would affect the constraints. Thank you. $\endgroup$
    – JessicaK
    Feb 5 '16 at 10:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.