Proving that a Hölder space is a Banach space

Question

I am trying to show that the Hölder space $C^{k,\gamma}(\bar{U})$ is a Banach space. To do this, I successfully proved that the mapping $\| \quad \| : C^{k,\gamma}(\bar{U}) \to [0,\infty)$ is a norm, by proving its properties.

But how do I show that the sequence $\{u_k\}_{k=1}^\infty \subset C^{k,\gamma}(\bar{U})$ converges to $u \in C^{k,\gamma}(\bar{U})$, that is, how do I show that $$\lim_{k \to \infty}\|u_k-u\|=0,$$ which would mean the normed linear space is complete, and hence a Banach space?

Here are the following taken from PDE Evans, 2nd edition, page 255:

Definition. The Hölder space $$C^{k,\gamma}(\bar{U})$$ consists of all functions $u \in C^k(\bar{U})$ for which the norm $$\|u\|_{C^{k,y}(\bar{U})}:= \sum_{|\alpha|\le k} \|D^\alpha u \|_{C(\bar{U})}+\sum_{|\alpha|=k} [D^\alpha u]_{C^{0,\gamma}(\bar{U})}$$ is finite.

Also from page 254,

Definitions. (i) If $u : U \to \mathbb{R}$ is bounded and continuous, we write $$\|u\|_{C(\bar{U})}:=\sup_{x\in U}|u(x)|.$$

(ii) The $\gamma$th-Hölder seminorm of $u : U \to \mathbb{R}$ is $$[u]_{C^{0,\gamma}(\bar{U})}:=\sup_{\substack{x,y\in U \\ x \neq y}} \left\{\frac{|u(x)-u(y)|}{|x-y|^\gamma} \right\},$$ and the $\gamma$th-Hölder norm is $$\|u\|_{C^{0,\gamma}(\bar{U})}:=\|u\|_{C(\bar{U})}+[u]_{C^{0,\gamma}(\bar{U})}.$$

This is all I got so far: \begin{align} \|u_k-u\|_{C^{k,\gamma}(\bar{U})}&=\sum_{|\alpha|\le k} \|D^\alpha u \|_{C(\bar{U})}+\sum_{|\alpha|=k} [D^\alpha u]_{C^{0,\gamma}(\bar{U})} \\ &= \sum_{|\alpha|\le k} \sup_{x\in U} |u_k(x)-u(x)|+ \sum_{|\alpha|=k} \sup_{\substack{x,y\in U \\ x \neq y}} \left\{\frac{|[u_k(x)-u(x)]-[u_k(y)-u(y)]|}{|x-y|^\gamma} \right\}. \end{align} Now, where can I go from here, to show that $\lim_{k \to \infty}\|u_k-u\|_{C^{k,\gamma}(\bar{U})}=0$? The sequence is Cauchy, and I have to use that fact somehow.

Answer 1

We have to show the following: given a Cauchy sequence $(u_n)_{n\in \mathbb N}$ in $C^{k,\gamma}(\overline U),$ there is a $u \in C^{k,\gamma}(\overline U)$ such that $\lim_{n\rightarrow\infty}u_n = u$ in $C^{k,\gamma}(\overline U).$

Note that the Hölder norm is the "sum" of the $C^k$ norm (i.e. sup-norm up to the $k$-th derivatives) and the Hölder condition with parameter $\gamma \in (0,1)$ for the $k$-th derivatives. From the $C^k$ part of the Hölder norm and the completeness of $C^k(\overline U),$ we get a $u \in C^k(\overline U)$ with $\lim_{n\rightarrow\infty}u_n = u$ in $C^k(\overline U).$ Now let's look at the Hölder condition. Let's fix a multi-index $\alpha$ with $|\alpha| = k$ and write $$ v := D^\alpha u\qquad and \qquad v_n := D^\alpha u_n $$ for short. Then we have for fixed $x\neq y$ $$ \begin{align} \frac{|v(x) - v(y)|}{|x-y|^\gamma} & = \frac{|v(x) - v_m(x) + v_m(x) - v_m(y) + v_m(y) - v(y)|}{|x-y|^\gamma} \\ & \leq \frac{|v(x) - v_m(x)|}{|x-y|^\gamma} + \frac{|v_m(x) - v_m(y)|}{|x-y|^\gamma} + \frac{|v_m(y) - v(y)|}{|x-y|^\gamma} \end{align} $$ for arbitrary $m \in \mathbb N.$ Here, the first and third summands on the r.h.s. can be made arbitrarily small by choosing $m$ large enough, since $v_m \rightarrow v$ uniformly, and the second summand on the r.h.s. is bounded independent of $m$ and $x\neq y$ since the sequence $u_n$ is Cauchy, hence bounded, in the Hölder norm. From all this we get a constant $M_\alpha$ such that $$ \frac{|v(x) - v(y)|}{|x-y|^\gamma} \leq M_\alpha \qquad independent\ of\ x\neq y. $$ This shows that we have $$ u\in C^{k,\gamma}(\overline U). $$ So far so good. Finally, we have to show that $u_n \rightarrow u$ in $C^{k,\gamma}(\overline U).$ From the construction of $u,$ we already have $u_n \rightarrow u$ in $C^k(\overline U),$ so we only have to show convergence in the Hölder seminorms of the $k$-th derivatives. We use the notation $\alpha,v,v_n$ from above. We have $$ \begin{align} \frac{|(v-v_m)(x) - (v-v_m)(y)|}{|x-y|^\gamma} & = \frac{|v(x) - v_m(x) - v(y) + v_m(y)|}{|x-y|^\gamma} \\ & = \frac{|(\lim_{k\rightarrow\infty}v_k(x)) - v_m(x) - (\lim_{k\rightarrow\infty}v_k(y)) + v_m(y)|}{|x-y|^\gamma} \\ & = \lim_{k\rightarrow\infty}\frac{|v_k(x) - v_m(x) - v_k(y) + v_m(y)|}{|x-y|^\gamma} \\ & = \lim_{k\rightarrow\infty}\frac{|(v_k - v_m)(x) - (v_k - v_m)(y)|}{|x-y|^\gamma}. \end{align} $$ Here, the r.h.s. can be made arbitrarily small independent of $x\neq y$ by choosing $m$ large enough, since the sequence $u_n$ is Cauchy in the Hölder norm. So we find $v_n \rightarrow v$ in the Hölder seminorm. Since there are only finitely many multi-indices $\alpha$ (or equivalently, $k$-th derivatives) to consider, we can conclude $$ u_n \rightarrow u\qquad in\ C^{k,\gamma}(\overline U), $$ as desired.

Answer 2

Since your comment to my first answer is tricky to read, let me restate it. The follow-up question is: how does $$ \frac{|v(x)-v(y)|}{|x-y|^\gamma} \le M_\alpha < \infty \qquad independent\ of\ x\neq y \qquad\qquad\qquad(*)$$ for every multi-index $\alpha$ of order $k$ imply $$ u \in C^{k,\gamma}(\bar{U})? $$ The answer is as follows: in order to show $u \in C^{k,\gamma}(\bar{U}),$ we have to show that all the ingredients of the Hölder norm of $u$ can be calculated, and the value of the Hölder norm thus obtained is finite.

By construction of $u,$ we have $$ \sum_{|\beta|\leq k} \|D^\beta u\|_{C(\overline U)} < \infty. $$ Moreover, (*) above implies that for a fixed multi-index $\alpha$ of order $k,$ we have $$ [D^\alpha u]_{C^{,\gamma}(\overline U)} \leq M_\alpha < \infty. $$ From this we get $$ \sum_{|\alpha| = k}[D^\alpha u]_{C^{,\gamma}(\overline U)} \leq \sum_{|\alpha| = k} M_\alpha < \infty. $$ So we have all in all $$ \|u\|_{C^{k,\gamma}(\bar{U})} = \sum_{|\beta|\leq k} \|D^\beta u\|_{C(\overline U)} + \sum_{|\alpha| = k}[D^\alpha u]_{C^{,\gamma}(\overline U)} < \infty $$ and thus $$ u \in C^{k,\gamma}(\bar{U}), $$ as desired.

Answer 3

The logic behind $v_n$ converging to $v$ in the Hölder seminorm sense is incorrect for the given answers below. Let me correct it:

We already know $v$ has finite Hölder seminorm $[v]$ and if $v_n$ is Cauchy in the Hölder seminorm, so will be $p_n = v_n - v$, so using norm inequality:

$$ \big|[p_n] - [p_m]\big|=\big|[v_n-v] - [v_m-v]\big| \le [(v_n-v)-(v_m-v)] < \varepsilon $$

for all $m,n > N$, for some $N$. So $[p_n] = [v_n - v]$ is a Cauchy sequence in $\mathbb{R}$, hence it will converge as $\mathbb{R}$ is complete.

Let's assume it converges to some nonzero positive value $r$. Let $\varepsilon$ be a positive number such that $r-\varepsilon > 0$.

We know:

\begin{align*} [v_n-v] &= \sup_{x\neq y} \frac{|v_n(x)-v(x)-(v_n(y)-v(y))|}{|x-y|^{\gamma}} \\ &\le \sup_{x\neq y} \frac{|v_n(x)-v(x)|}{|x-y|^{\gamma}} + \sup_{x\neq y} \frac{|v_n(y)-v(y)|}{|x-y|^{\gamma}} . \tag{1} \end{align*}

Now as $[v_n - v]$ converges to $r$ , for the ɛ, there is a N such that $[v_n - v]$ > r - ɛ for all n > N. So for each n>N, there is at least one $(x,y)$ with $x\neq y$ such that:

$$ \frac{|v_n(x)-v(x)-(v_n(y)-v(y))|}{|x-y|^{\gamma}} > r-\varepsilon.\tag{2} $$

Now again: As $v_m$ converges in the sup norm hence uniformly to $v$, we can choose another $N'_{xy}$ (which is (x,y) dependent) such that for all $m > N'_{xy}$:

$$ \frac{|v_m(x)-v(x)|}{|x-y|^{\gamma}} < \frac{r-\varepsilon}{4} $$

and

$$ \frac{|v_m(y)-v(y)|}{|x-y|^{\gamma}} < \frac{r-\varepsilon}{4}, $$

so that using equation (1):

$$ \frac{|v_m(x)-v(x)-(v_m(y)-v(y))|}{|x-y|^{\gamma}} < \frac{r-\varepsilon}{2}.\tag{3} $$

for all $m > N_{xy}'$ and for the (x,y), x≠y.

Now $v_n$ being cauchy, there is a M such that for n,m>M, [$v_n-v_m$]<(r-ɛ)/2 so for all n , m > M:

\begin{align*} \frac{|v_n(x)-v(x)-(v_n(y)-v(y))|}{|x-y|^{\gamma}} &= \frac{|v_n(x)-v_m(x)-(v_n(y)-v_m(y))+v_m(x)-v(x)-(v_m(y)-v(y))|}{|x-y|^{\gamma}}\\ &\le \frac{|v_n(x)-v_m(x)-(v_n(y)-v_m(y))|}{|x -y|^{\gamma}}+\frac{|v_m(x)-v(x)-(v_m(y)-v(y))|}{|x-y|^{\gamma}}\\ &\le [v_n-v_m]+\frac{|v_m(x)-v(x)-(v_m(y)-v(y))|}{|x-y|^{\gamma}} \\ &\le (r-ɛ)/2 + \frac{|v_m(x)-v(x)-(v_m(y)-v(y))|}{|x-y|^{\gamma}}\tag{4} \end{align*} So for any n>max(M,N), for any (x,y),x≠y, we can take m>max(M,$N_{xy}$) so that,using equation (3) and (4), we will have: $$ \frac{|v_n(x)-v(x)-(v_n(y)-v(y))|}{|x-y|^{\gamma}} < {r-\varepsilon}.\tag{5} $$ for all n>max(M,N) for each (x,y),x≠y.

But clearly equation (2) and (5) pose a contradiction. Because for any n>max(M,N), we should be able to find a (x,y),x≠y such that (2) is true. Therefore, $[v_n - v]$ must converge to $r=0$.
Hence $v_n$ converges to $v$ in the Hölder seminorm sense too.

Answer 4

There seems to be some confusion about how the triangle inequality is applied in the accepted answer, so I address that here as well as provide an alternative proof. First, for a given $\varepsilon > 0$, fix $N$ large enough such that $$\tag{$\star$}\|u_{n}-u_{m}\|_{C^{k, \gamma}(\overline{U})}\leq \varepsilon\text{ for }n, m\geq N.$$ Denote $v_{n}:= D^{\alpha}u_{n}$ and $v = D^{\alpha}u$ for some fixed multi-index $\alpha$ with $|\alpha| = k$. Then fixing $x, y\in U$ with $x\neq y$ and appplying the triangle inequality with $m\geq N$, we have $$\frac{|v(x)-v(y)|}{|x-y|^{\gamma}}\leq \frac{|v(x)-v_{m}(x)|}{|x-y|^{\gamma}}+\frac{|v_{m}(x)-v_{m}(y)|}{|x-y|^{\gamma}}+\frac{|v_{m}(y)-v(y)|}{|x-y|^{\gamma}}.$$ The equation $(\star)$ in particular implies that $\|v_{n}-v_{m}\|_{C(\overline{U})}\leq \varepsilon$. Moreover, since $\{u_{n}\}$ is Cauchy in $\|\cdot\|_{C^{k, \gamma}(\overline{U})}$, we have $\sup_{n}[v_{n}]_{C^{0, \gamma}(\overline{U})}\leq M$ for some constant $M > 0$. Using these facts, we get $$\frac{|v(x)-v(y)|}{|x-y|^{\gamma}}\leq\frac{2\varepsilon}{|x-y|^{\gamma}}+M.$$ Since $\varepsilon > 0$ is arbitrary, we may take the limit as $\varepsilon\rightarrow 0$ (still for fixed $x\neq y$) to obtain $\frac{|v(x)-v(y)|}{|x-y|^{\gamma}}\leq M$. Finally, since $x\neq y$ are arbitrary, we conclude that $[v]_{C^{0, \gamma}(\overline{U})}\leq M$.

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Here's an alternative proof. Fix $x\neq y$ as above. Using $(\star)$, we have $$\frac{|(v_{n}-v_{N})(x)-(v_{n}-v_{N})(y)|}{|x-y|^{\gamma}}\leq\varepsilon\text{ for }n\geq N.$$ Taking the limit as $n\rightarrow\infty$ we get $$\frac{|(v-v_{N})(x)-(v-v_{N})(y)|}{|x-y|^{\gamma}}\leq\varepsilon$$ and thus $$\frac{|v(x)-v(y)|}{|x-y|^{\gamma}}\leq \varepsilon+[v_{N}]_{C^{0, \gamma}(\overline{U})}.$$