# Fundamental Theorem of Calculus for distributions.

Consider a function $F \in C^{\alpha}( \mathbb{R})$ for $0 < \alpha < 1.$ Then we can take it's distributional derivative. We can say $f = F' \in C^{\alpha -1}( \mathbb{R} )$. My issue is going back.

Say I have a $\alpha -1 -$Hölder function $f$, then how can I "integrate" it to get a primitive $F$ such that $f$ is it's distributional derivative?

Here we use the following definition (but any other definition can be used as well) for negative $\beta \in (-1,0)$: $$g \in C^{\beta}( \mathbb{R} ) \Leftrightarrow \ |\langle g, \phi_{x}^{\lambda} \rangle| \ ≤ C \lambda^{\beta}$$ where $\phi_{x}^{\lambda}(z) = \lambda^{-d}\phi(\frac{z -x}{\lambda})$ and $C$ is uniform over all $x$ and $\phi \in C^{\infty}_0(-1,1)$ with $||\phi||_{C^1}≤1.$

EDIT:

1. An approach would be to prove that $$C^{\beta} \subseteq L^1_{Loc}$$ which would mean in particular that our distributions are actually functions. This does not work, as pointed out in the comments, and in related questions.

2. My other idea was to define $\langle f, 1_{[0,a]} \rangle$ by using typical approximation of the indicator function of $1_{[0,a]}.$ This cannot work: for example the derivative of a the Brownian motion is not a measure.

I saw that this topic is covered only slightly in MathStackExchange. Here are similar questions regarding this topic.

1. Here Is a general question about Holder spaces with negative exponents.
2. Here Is a question that is very similar to mine, maybe just in a slightly different context.
3. Here Is a question by myself regarding the same topic. In fact as you might imagine I am looking into this subject with little success :D

Finally let me also give some motivation as to why I ma studying this: these spaces are used in the theory of SPDEs. In particular a reference is Fritz's and Hairer's book: "A course on rough paths." Both my questions are exercises in this book.

• We don't have $C^{\beta} \subseteq L^1_{\text{loc}}$ for $\beta < 0$. The Cantor function is $\dfrac{\log 2}{\log 3}$-Hölder continuous, and its distributional derivative is a measure that is singular with respect to the Lebesgue measure (it's concentrated on the Cantor set). I see that this was also given as an example in the answer to the second question, and there were also given examples of Hölder-continuous functions that are not of bounded variation, hence their derivatives aren't even measures. Feb 17, 2017 at 21:04
• @DanielFischer Thanks for the comment. That seems to settle things. Although it might seem awkward that I ask a question answered in a linked question, it's just because I found the latter question later :) In any case maybe $C^{\beta}$ can be embedded in a space of measures? This seems to be exactly what I need. Feb 17, 2017 at 21:23
• The answer at the other question only answers part of yours (the derivative of a Hölder-continuous function is in general not a nice function). Since your question is an exercise in a book, I guess there is some reasonable way to construct $F$ (or maybe a reasonable distribution in $C^{\alpha-1}$ that is not the derivative of a $C^{\alpha}$ function, depending on whether the exercise says one can, or asks whether one can go back). But at the moment, I have no idea how to go about it. Feb 17, 2017 at 21:30
• In the book the question is supposed to receive a positive answer. In fact I used this result later in another exercise. At some other point in the book a similar problem appears. Here the authors suggest an approximation argument. Feb 17, 2017 at 22:34

I don't have the book you mention, so I don't know what tools you are allowed to use to solve those exercises, but anyway. First you need to show that your distribution $g$ has order one. Take $\psi\in C_{c}^{\infty}(\mathbb{R})$ with support in some interval $K=[-n,n]$ and $\Vert\psi\Vert_{C^{1}}\leq1$. Then $\phi(x):=\psi(nx)$, $x\in\mathbb{R}$, has support in $[-1,1]$ and so by applying the bound with $\lambda=n$ and $x=0$ you get that $\phi_{0}% ^{n}(x)=n^{-1}\phi(x/n)=n^{-1}\psi(x)$. Hence, by linearity, $$|\langle g,\psi\rangle|=n|\langle g,n^{-1}\psi\rangle|=n|\langle g,\phi _{0}^{n}\rangle|\leq n(Cn^{\beta}).$$ Now for every $\psi\in C_{c}^{\infty}(\mathbb{R})$ with support in $K$, applying the previous inequality with $\psi$ replaced by $\psi/\Vert\psi \Vert_{C^{1}}$, you get, again by linearity, $$|\langle g,\psi\rangle|\leq Cn^{1+\beta}\Vert\psi\Vert_{C^{1}}=C_{K}\Vert \psi\Vert_{C^{1}}.$$ This shows that $g$ has order one. Now any distribution can be written as sum of derivatives of continuous functions and if the order is $m$ you only need $m+2$ functions (you can find this in Edward's functional, or Rudin's functional analysis, or Leoni's Sobolev books).
So in your case you can find $f_{0}$, $f_{1}$, $f_{2}$ and $f_{3}$ continuous such that $$\langle g,\psi\rangle=\int_{\mathbb{R}}f_{0}\psi-\int_{\mathbb{R}}f_{1}% \psi^{\prime}+\int_{\mathbb{R}}f_{2}\psi^{\prime\prime}-\int_{\mathbb{R}}% f_{3}\psi^{\prime\prime\prime}.$$ Define $F_i(x):=\int_{0}^{x}f_i(t)\,dt$ for $i=0,1,2$ and $$\langle G,\psi\rangle=\int_{\mathbb{R}}F_{0}\psi-\int_{\mathbb{R}}F_{1}% \psi^{\prime}+\int_{\mathbb{R}}F_{2}\psi^{\prime\prime}-\int_{\mathbb{R}}% F_{3}\psi^{\prime\prime\prime}.$$ Then integrating by parts \begin{align}\langle G^\prime,\psi\rangle=-\langle G,\psi^\prime\rangle &=-\int_{\mathbb{R}}F_{0}\psi^\prime+\int_{\mathbb{R}}F_{1}% \psi^{\prime\prime}-\int_{\mathbb{R}}F_{2}\psi^{\prime\prime\prime}+\int_{\mathbb{R}}% F_{3}\psi^{\prime\prime\prime\prime}\\ &=\int_{\mathbb{R}}f_{0}\psi-\int_{\mathbb{R}}f_{1}% \psi^{\prime}+\int_{\mathbb{R}}f_{2}\psi^{\prime\prime}-\int_{\mathbb{R}}% f_{3}\psi^{\prime\prime\prime} \end{align}
There are probably simpler proofs done by hand (probably the proof that a distribution can be written as sum of derivatives has all the tricks you need to find a primitive). Mollifying $g$ might also do the trick.
• Thank you for the answer, and sorry for commenting so late. I understand your proof of the fact that $g$ is of order one. But with your definition shouldn't we have: $G' = \sum f_i$? I don't see why this is equal to $g$. Feb 23, 2017 at 6:46