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Fall 2015 - Problem 5

Baire category theorem

A function $f \in C\p{\br{0,1}}$ is called Hölder continuous of order $\delta > 0$ if there is a constant $C > 0$ such that $\abs{f\p{x} - f\p{y}} \leq C\abs{x - y}^\delta$ , $x, y \in \br{0, 1}$ . Show that the Hölder continuous functions form a set of the first category (a meager set) in $C\p{\br{0,1}}$ .

Solution.

We denote the space of Hölder continuous functions on $\br{0, 1}$ of order $\delta > 0$ via $H^\delta$ .

First, let us show that for any $\delta > 0$ , $H^\delta$ is nowhere dense. It suffices to show that for any $f \in C\p{\br{0,1}}$ and $\epsilon > 0$ , there exists $g \in C\p{\br{0,1}} \setminus H^\delta$ such that $\norm{f - g}_{L^\infty} < \epsilon$ . Indeed, this implies that any open neighborhood in $H^\delta$ contains an element outside of $H^\delta$ , i.e., $H^\delta$ has empty interior.

Recall that piecewise linear continuous functions are dense in $C\p{\br{0,1}}$ (e.g., $f$ is uniformly continuous, so we can approximate $f$ uniformly with a piecewise linear function on a fine enough partition of $\br{0, 1}$ ). Thus, we may assume without loss of generality that $f$ is a piecewise linear function. In particular, $f$ is some linear function $f\p{0} + mx$ on $\br{0, \eta}$ for some $\eta > 0$ .

Our prototype for functions in $H^\alpha \setminus H^\beta$ are $x^\alpha$ where $\alpha < \beta$ . Indeed, in this case, Jensen's inequality applied to the concave function $t \mapsto t^\beta$ yields

\abs{x^\alpha - y^\alpha} = \abs{\int_0^1 t^\alpha\chi_{\br{x,y}} \,\diff{t}} \leq \abs{\p{\int_0^1 t\chi_{\br{x,y}} \,\diff{t}}^\alpha} = \abs{x - y}^\alpha,

but

\frac{\abs{x^\alpha - y^\alpha}}{\abs{x - y}^\beta} = \frac{\abs{x^\alpha - y^\alpha}}{\abs{x - y}^\alpha} \cdot \frac{1}{\abs{x - y}^{\beta-\alpha}} \xrightarrow{\abs{x-y} \to 0} \infty,

since the first term is bounded. Thus, our goal is to approximate $f$ near $0$ with $x^{\delta/2} \notin H^\delta$ . If $a > 0$ , then

0 \leq x^{\delta/2} - x \leq a^{\delta/2}

on $\br{0, a}$ , since $x \in \br{0, 1}$ , so $x^{\delta/2} \geq x$ , and because $x^{\delta/2}$ is an increasing function. Let $a \in \br{0, t}$ be small enough such that $a^{\delta/2} < \epsilon$ . Then define

h\p{x} = \begin{cases} x^{\delta/2} & \text{if } 0 \leq x \leq a, \\ a^{\delta/2} & \text{if } a \leq x \leq a^{\delta/2}, \\ x & \text{if } a^{\delta/2} \leq x \leq t, \end{cases}

which is a continuous by construction. Essentially, $h$ is $x^{\delta/2}$ for a small amount of time before becoming $x$ . We claim that

g\p{x} = \begin{cases} f\p{0} + mh\p{x} & \text{if } 0 \leq x \leq t, \\ f\p{x} & \text{if } t \leq x \leq 1. \end{cases}

has the desired properties. On $\br{0, a}$ , we get

\abs{f\p{x} - g\p{x}} = \abs{\p{f\p{0} + mx} - \p{f\p{0} + mx^{\delta/2}}} = m\abs{x^{\delta/2} - x} \leq m\epsilon.

On $\br{a, a^{\delta/2}}$ ,

\begin{aligned} \abs{f\p{x} - g\p{x}} = \abs{\p{f\p{0} + mx} - \p{f\p{0} + ma^{\delta/2}}} &= m\abs{x - a^{\delta/2}} \\ &\leq m\abs{a - a^{\delta/2}} \\ &< m\epsilon. \end{aligned}

Finally, on $\br{a^{\delta/2}, 1}$ , $g\p{x} = f\p{x}$ by construction, so $\norm{f - g}_{L^\infty} \leq m\epsilon$ , which proves that $H^\delta$ is nowhere dense.

Now, define

E_{n,m} = \set{f \in C\p{\br{0,1}} \st \abs{f\p{x} - f\p{y}} \leq n\abs{x - y}^{1/m}}

so that

H^{1/m} = \bigcup_{n=1}^\infty E_{n,m}.

Notice that $E_{n,m}$ is closed in $C\p{\br{0,1}}$ : if $\set{f_n}_n \subseteq E_{n,m}$ is a sequence which converges uniformly to $f \in C\p{\br{0,1}}$ , then $\abs{f_k\p{x} - f_k\p{y}} \leq n\abs{x - y}^{1/m}$ for all $k \geq 1$ . Sending $k \to \infty$ , we easily see that $f \in E_{n,m}$ as well. Finally, observe that if $\alpha \leq \beta$ , then $\abs{x - y} \leq 1 \implies \abs{x - y}^\alpha \leq \abs{x - y}^\beta$ . Hence, $H^\alpha \subseteq H^\beta$ , which means the Hölder continuous functions are given by

\bigcup_{m=1}^n H^{1/m} = \bigcup_{m=1}^n \bigcup_{n=1}^\infty E_{n,m},

so it is a countable union of closed, nowhere dense sets, hence meager.