Section 3.2 Hölder Spaces
Definition 3.3.
Let \(\alpha \in (0, 1]\text{.}\) Then \(C^{0, \alpha}([a,b])\) is the subset of \(C^0([a,b])\) comprising all \(f \in C^0([a,b])\) such that the following quantity is finite:
\begin{equation}
[f]_{C^{0, \alpha}([a,b])} = \sup_{x, y \in [a,b], \ x \ne y} \frac{|f(x) - f(y)|}{|x - y|^\alpha} \lt \infty.\tag{3.1}
\end{equation}
For \(f \in C^{0, \alpha}([a,b])\text{,}\) we define
\begin{equation*}
\n f_{C^{0, \alpha}([a,b])} = \n f_{C^0([a,b])} + [f]_{C^{0, \alpha}([a,b])}.
\end{equation*}
Remark 3.4.
- A function satisfying (3.1) is called Hölder continuous. For \(\alpha = 1\text{,}\) the condition is equivalent to Lipschitz continuity, and \(L=[f]_{C^{0,1}([a,b])}\) is the smallest Lipschitz constant for for \(f\text{.}\)
- One can check that \([\blank]_{C^{0, \alpha}([a,b])}\) is not a norm on \(C^{0,\alpha}([a,b])\text{.}\) It is called a Hölder seminorm. It can be equivalently defined as the smallest constant \(M \ge 0\) for which \(\abs{f(x)-f(y)} \le M\abs{x-y}^\alpha\) for all \(x,y \in [a,b]\text{.}\)
Theorem 3.5.
Let \(\alpha \in (0, 1]\text{.}\) Then \(C^{0, \alpha}([a,b])\) is a Banach space.
Proof.
The vector space and norm properties are routine, so we concentrate on the completeness. Let \((f_n)_{n \in \N}\) be a Cauchy sequence in \(C^{0, \alpha}([a,b])\text{.}\) Then it is also Cauchy in \(C^0([a,b])\text{.}\) Hence there exists a uniform limit \(f = \lim_{n \to \infty} f_n\text{.}\) We need to show that \(f \in C^{0, \alpha}([a,b])\) and that the convergence holds in \(C^{0, \alpha}([a,b])\text{.}\)
By Exercise 1.3.3, there exists \(M \gt 0\) such that \(\|f_n\|_{C^{0, \alpha}([a,b])} \le M\) for all \(n \in \N\text{.}\) Hence for any \(x, y \in [a,b]\) with \(x \ne y\text{,}\)
\begin{equation*}
\frac{|f(x) - f(y)|}{|x - y|^\alpha} = \lim_{n \to \infty} \frac{|f_n(x) - f_n(y)|}{|x - y|^\alpha} \le M.
\end{equation*}
In particular, it follows that
\begin{equation*}
[f]_{C^{0, \alpha}([a,b])} \lt \infty,
\end{equation*}
and therefore \(f \in C^{0, \alpha}([a,b])\text{.}\)
In order to prove the convergence, fix \(\varepsilon \gt 0\) and choose \(N \in \N\) such that \(\|f_m - f_n\|_{C^{0, \alpha}([a,b])} \lt \varepsilon\) whenever \(m, n \ge N\text{.}\) Then for any \(x, y \in [a,b]\) with \(x \ne y\text{,}\)
\begin{align*}
\amp\frac{|(f(x) - f_n(x)) - (f(y) - f_n(y))|}{|x - y|^\alpha} \\
\amp= \lim_{m \to \infty} \frac{|(f_m(x) - f_n(x)) - (f_m(y) - f_n(y))|}{|x - y|^\alpha} \\
\amp\le \limsup_{m \to \infty} [f_m - f_n]_{C^{0, \alpha}([a,b])} \le \varepsilon,
\end{align*}
provided that \(n \ge N\text{.}\) Hence \([f - f_n]_{C^{0, \alpha}([a,b])} \le \varepsilon\text{,}\) and since \(\varepsilon\) was chosen arbitrarily and we already have uniform convergence, this shows that \(f_n \to f\) in \(C^{0, \alpha}([a,b])\text{.}\)
Theorem 3.6.
\(C^1([a,b])\) is a closed normed subspace of \(C^{0, 1}([a,b])\text{.}\)
Proof.
You are asked to prove this in Exercise 3.2.3.
Exercises Exercises
1. Hölder continuity in metric spaces.
Let \((X, d_X)\) and \((Y, d_Y)\) be metric spaces and let \(\alpha \in (0, 1]\text{.}\) If \(f \maps X \to Y\) is a map such that there exists \(L \ge 0\) satisfying the inequality
\begin{equation*}
d_Y(f(x), f(y)) \le L \left(d_X(x, y)\right)^\alpha,
\end{equation*}
then we say that \(f\) is Hölder continuous (or Lipschitz continuous if \(\alpha = 1\)). Show that any Hölder (or Lipschitz) continuous map is uniformly continuous.
2. (PS7) Regularity of \(x \mapsto x^\alpha\).
Let \(\alpha \in (0,1)\) and consider the function \(f \maps [0,1] \to
\R\) defined by \(f(x) = x^\alpha\text{.}\)
(a)
Show that \(f \notin C^1([0,1])\text{.}\)
Hint.
Show that there is no \(g \in C^0([0,1])\) with \(g(t)=f'(t)\) for \(t \in (0,1)\text{.}\)
(b)
Prove that \(f \notin C^{0,\beta}([0,1])\) for \(\beta \in (\alpha,1)\text{.}\)
Hint.
Estimate the supremum in (3.1) from below by setting \(y=0\text{.}\)
(c)
Show that \(f \in C^{0,\alpha}([0,1])\text{.}\)
Hint.
First, use calculus to show that
\begin{gather}
(t+1)^\alpha-t^\alpha \le 1 \text{ for } t \ge 0\text{.}\tag{✶}
\end{gather}
\begin{equation*}
\frac{x^\alpha-y^\alpha}{(x-y)^\alpha} \le 1\text{.}
\end{equation*}
3. (PS7) \(C^1\) as a subset of \(C^{0,1}\).
Let \(a, b \in \R\) with \(a \lt b\text{.}\)
(a)
Show that \(C^1([a,b])\) is a linear subspace of \(C^{0, 1}([a,b])\text{.}\)
Hint.
As you can easily check yourself, \(C^1([a,b])\) is closed under vector space operations, and so the main thing to prove here is that it is a subset of \(C^{0,1}([a,b])\text{.}\) To prove this, fix \(x,y \in [a,b]\) with \(x \ne y\text{,}\) and try to estimate the quotient in the definition of \([f]_{C^{0,1}([a,b])}\) in terms of \(f'\) using the mean value theorem. Taking a supremum, conclude that \([f]_{C^{0,1}([a,b])} \le \n
{f'}_{C^0([a,b])}\text{.}\)
(b)
Show that, for \(f \in C^1([a,b])\text{,}\) \(\n{f'}_{C^0([a,b])} = [f]_{C^{0,1}([a,b])}\text{.}\)
Hint.
In the previous part you have hopefully already shown that \(\n{f'}_{C^0([a,b])} \ge [f]_{C^{0,1}([a,b])}\) for any \(f \in
C^1([a,b])\text{,}\) and so it suffices to show the reverse inequality. Fix \(x \in (a,b)\text{,}\) and write \(\abs{f'(x)}\) as a limit as \(y \to x\text{,}\) and then estimate inside the limit.
(c)
Show that, for \(f \in C^1([a,b])\text{,}\) \(\n f_{C^{0,1}([a,b])}= \n f_{C^1([a,b])}\text{.}\) Conclude that \(C^1([a,b])\) (equipped with \(\n\blank_{C^1([a,b])}\)) is a normed subspace of \(C^{0,1}([a,b])\) (equipped with \(\n\blank_{C^{0,1}([a,b])}\)).
(d)
Show that \(C^1([a,b])\) is closed in \(C^{0, 1}([a,b])\text{.}\)
Hint.
Part c is useful here, as is the completeness of \(C^1([a,b])\) when equipped with the usual \(\n\blank_{C^1([a,b])}\) norm.
4. \(C^{0,\alpha}\) is non-separable.
Let \(\alpha \in (0,1)\) and \(a,b \in \R\) with \(a \lt b\text{,}\) and consider the normed space \(C^{0,\alpha}([a,b])\) of Hölder continuous functions (Definition 3.3). For every \(t \in [a,b]\text{,}\) define a function \(f_t
\in C^{0,\alpha}([a,b])\) by \(f_t(x) = \abs{x-t}^\alpha\text{.}\)
(a)
Show that, if \(s,t \in [a,b]\) and \(s \ne t\text{,}\) then
\begin{equation*}
[f_t-f_s]_{C^{0,\alpha}([a,b])}
\ge \lim_{x \to t}
\frac{\abs{\abs{x-t}^\alpha-\abs{x-s}^\alpha+\abs{t-s}^\alpha}}{\abs{x-t}^\alpha}
= 1\text{.}
\end{equation*}
Conclude that \(\n{f_t-f_s}_{C^{0,\alpha}([a,b])} \ge 1\text{.}\)
Hint.
This is probably the hardest part of this problem. If you are getting stuck, I would recommend moving on to the others parts and doing them first. One way to compute the limit is to compare some of the terms to the limit definition of the derivative \(f_s'(t)\text{.}\)
(b)
Suppose that \(G \subseteq C^{0,\alpha}([a,b])\) is dense. Show that for each \(t \in [a,b]\) there must exist \(g_t \in G\) with \(\n{g_t-f_t}_{C^{0,\alpha}([a,b])} \lt 1/2\text{.}\)
(c)
(d)
Conclude that \(G\) cannot be countable, and hence that \(C^{0,\alpha}([a,b])\) is not separable (Definition 2.5).
5. (PS8) \(C^{1,\alpha}([a,b])\).
For \(\alpha \in (0,1]\text{,}\) the set \(C^{1,\alpha}([a,b])\) comprises all \(f \in
C^1([a,b])\) such that (the continuous extension of) \(f' \in
C^{0,\alpha}([a,b])\text{.}\) Show that the norm
\begin{equation}
\n f_{C^{1,\alpha}([a,b])} =
\n f_{C^1([a,b])} + [f']_{C^{0,\alpha}([a,b])}\tag{3.2}
\end{equation}
makes \(C^{1,\alpha}([a,b])\) a Banach space. You do not have to check that it is a normed space.
Hint 1.
Theorem 3.2 and Theorem 3.5 are very useful here. You might also want to look at their proofs for some general inspiration.
Hint 2.
Suppose that \(\seq fn\) is Cauchy in \(C^{1,\alpha}([a,b])\text{.}\) Argue that \(\seq fn\) is also Cauchy in \(C^1([a,b])\) and that \(\seq{f'}n\) is Cauchy in \(C^{0,\alpha}([a,b])\text{.}\) Now appeal to the completeness of \(C^1([a,b])\) and \(C^{0,\alpha}([a,b])\text{.}\)
Hint 3.
Suppose that \((X,d_X)\) and \((Y,d_Y)\) are metric spaces with \(Y \subset
X\text{.}\) Then a sequence \(\seq yn\) in \(Y\) is also a sequence in \(X\text{.}\) Without knowing how the metrics \(d_X,d_Y\) compare, however, we can conclude nothing about how \(\seq yn\) being Cauchy/convergent as a sequence in \((Y,d_Y)\) is related to it being Cauchy/convergent as a sequence in \((X,d_X)\text{.}\)
6. (PS7) Inclusions between Hölder spaces.
Let \(0 \lt \alpha \lt \beta \le 1\) and \(f \in C^{0,\beta}([a,b])\text{.}\) Show that
\begin{equation*}
[f]_{C^{0,\alpha}([a,b])} \le \abs{b-a}^{\beta-\alpha} [f]_{C^{0,\beta}([a,b])} \text{,}
\end{equation*}
and conclude that \(C^{0,\beta}([a,b]) \subseteq C^{0,\alpha}([a,b])\text{.}\)
Hint.
Write \(\abs{x-y}^\beta=\abs{x-y}^{\beta-\alpha} \abs{x-y}^\alpha\text{.}\)
