Characterization of $C^{k,\alpha}$ (functions with Hölder continuous derivatives) through Taylor estimates

Solution 1:

We have $$ |f(x+h)-P_{x}(h)|\leq c|h|^{k+\alpha}% $$ for all $x\in\mathbb{R}$ and all $h$ with $|h|\leq1$. Write $$ P_{x}(h)=\sum_{n=0}^{k}\frac{1}{n!}a_{n}(x)h^{n}. $$ Taking $h=0$ you get $f(x)=a_{0}(x)$ so $$ \left\vert f(x+h)-f(x)-\sum_{n=1}^{k}\frac{1}{n!}a_{n}(x)h^{n}\right\vert \leq c|h|^{k+\alpha}. $$ Next, by the triangle inequality \begin{align*} \left\vert f(x+h)-f(x)-a_{1}(x)h\right\vert & \leq\left\vert f(x+h)-f(x)-\sum _{n=1}^{k}\frac{1}{n!}a_{n}(x)h^{n}\right\vert +\sum_{n=2}^{k}|a_{n}% (x)||h|^{n}\\ & \leq c|h|^{k+\alpha}+\sum_{n=2}^{k}\frac{1}{n!}|a_{n}(x)||h|^{n}% \end{align*} and so% $$ \left\vert \frac{f(x+h)-f(x)-a_{1}(x)h}{h}\right\vert \leq c|h|^{k-1+\alpha }+\sum_{n=2}^{k}\frac{1}{n!}|a_{n}(x)||h|^{n-1}\rightarrow0 $$ as $h\rightarrow0$, which shows that $f$ is differentiable with $f^{\prime }(x)=a_{1}(x)$.

%%%%%This part feels wrong but I cannot find the mistake%%%%%%%%%%%% Hence, for $k=1$, $$ \left\vert f(x+h)-f(x)-f^{\prime}(x)h\right\vert \leq c|h|^{1+\alpha}. $$

In turn, replacing $x$ with $x-h$ we get $$ \left\vert f(x)-f(x-h)-f^{\prime}(x-h)h\right\vert \leq c|h|^{1+\alpha}% $$ while replacing $h$ with $-h$ gives $$ \left\vert f(x-h)-f(x)+f^{\prime}(x)h\right\vert \leq c|h|^{1+\alpha}. $$ Now \begin{align*} |f^{\prime}(x)-f^{\prime}(x-h)| & \leq\left\vert \frac{f(x-h)-f(x)} {h}+f^{\prime}(x)\right\vert +\left\vert -\frac{f(x-h)-f(x)}{h}-f^{\prime }(x-h)\right\vert \\ & =\left\vert \frac{f(x-h)-f(x)}{h}+f^{\prime}(x)\right\vert +\left\vert \frac{f(x)-f(x-h)}{h}-f^{\prime}(x-h)\right\vert \\ & \leq2c|h|^{\alpha}. \end{align*}

This proves that $f^{\prime}$ is Holder's continuous.

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For $k\geq2$ we first prove that all the functions $a_{n}$ are bounded in $[a,b]$. We will use the following fact about polynomials. Given $k\in\mathbb{N}$, let $V$ be the vector space of all polynomials $P:[-1,1]\rightarrow\mathbb{R}$ of degree less than or equal to $k$. Given $P\in V$, let $\Vert P\Vert:=\max\{|a_{0}|,\ldots,|a_{k}|\}$, where $P(t)=a_{0}+\cdots+a_{k}t^{k}$, $t\in\lbrack-1,1]$. Then $\Vert\cdot\Vert$ is a norm in $V$.

Since the vector space $V$ has finite dimension $k$, all norms are equivalent. In particular,% $$ c_{1}\Vert P\Vert_{L^{\infty}([-1,1])}\leq\Vert P\Vert\leq c_{2}\Vert P\Vert_{L^{\infty}([-1,1])}% $$ for all $P\in V$ and for some constants $c_{1}>0$ and $c_{2}>0$.

Now let $P\in V$ be such that $\Vert P\Vert_{L^{\infty}([-1,1])}\leq L$. Then by the previous inequality,% $$ \Vert P\Vert:=\max\{|a_{0}|,\ldots,|a_{k}|\}\leq c_{2}\Vert P\Vert_{L^{\infty }([-1,1])}\leq c_{2}L, $$ which implies that $|a_{n}|\leq c_{2}L$ for all $n=1, \ldots, k$.

Since $f$ is continuous, there exists $$ M=\max_{[a-1,b+1]}|f(x)|<\infty $$ and so for $-1\leq h\leq1$ we have \begin{align*} \left\vert \sum_{n=1}^{k}\frac{1}{n!}a_{n}(x)h^{n}\right\vert & \leq\left\vert f(x+h)-f(x)-\sum_{n=1}^{k}\frac{1}{n!}a_{n}(x)h^{n}\right\vert +\left\vert f(x+h)-f(x)\right\vert \\ & \leq c|h|^{k+\alpha}+2M\leq c+2M. \end{align*} Hence, $$ \left\vert \sum_{n=1}^{k}\frac{1}{n!}a_{n}(x)h^{n}\right\vert \leq c+2M $$ for all $h\in\lbrack-1,1]$. It follows from the property about polynomials above that there exists a constant $C=C(a,b,M,c)>0$ such that% $$ |a_{n}(x)|\leq C $$ for all $x\in\lbrack a,b]$ and all $n=1,\ldots,k$.

We can now apply Theorem 3 in Oliver's paper to conclude that $a_{n}% (x)=f^{(n)}(x)$.