Let $f$ be a compactly supported $C^{\alpha}$ function (that is Holder continuous with exponent $\alpha$) and let $g$ be a compactly supported $C^\beta$ function. What can we say about Holder continuity of their convolution $$ h(x):=\int f(z-x) g (z) dz? $$

It is quite clear that $h$ is in class $C^{\max(\alpha,\beta)}$, but is it possible to say something better? Is it true that $h$ is in class $C^{\alpha+\beta}$? How one can prove something like this?

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    $\begingroup$ At least not without restrictions on $\alpha$ and $\beta$. We cannot expect the convolution of a Lipschitz ($\alpha = 1$) function with any other Holder class to be any better than Lipschitz; otherwise the convolution would be constant. $\endgroup$
    – Eric Thoma
    Commented Mar 23, 2016 at 20:02
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    $\begingroup$ @EricThoma Could you please explain your point? If $\alpha=1$ and $\beta=1$ why their convolution cannot be from the class $C^2$ (that is twice continuously differentiable)? $\endgroup$
    – Oleg
    Commented Mar 27, 2016 at 12:13
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    $\begingroup$ Sure. Functions that are non-constant are not Holder with exponent $\alpha$ for any $\alpha > 1$. The class $C^2$ of twice continuously differentiable functions is not the same thing as Holder class with exponent $2$; you are overloading the $C^\alpha$ notation. $\endgroup$
    – Eric Thoma
    Commented Mar 27, 2016 at 16:36
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    $\begingroup$ @EricThoma, thanks for your comments. I used a standard notation. $C^{n+\alpha}$ where $n$ is an integer and $\alpha\in[0,1)$ is the class of all functions that have $n$ derivatives and whose $n$-th derivative is a Holder function with exponent $\alpha$. $\endgroup$
    – Oleg
    Commented Mar 29, 2016 at 14:13
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    $\begingroup$ Anyway, let's sat that $\alpha$ and $\beta$ are small. Does the statement hold for such $\alpha$ and $\beta$? $\endgroup$
    – Oleg
    Commented Mar 29, 2016 at 14:15

2 Answers 2


The statement is actually true. I guess it is possible to prove in many different ways. For the purpose of generality, and because it is the only proof that I came up with by chance, my proof will refer to Hölder-Besov spaces over a torus $\mathbb(T)$.

Let us suppose that $f \in \mathcal{C}^{\alpha} = \mathcal{B}^{\alpha}_{\infty , \infty}$ and $g \in \mathcal{C}^{\beta}$ with $\alpha, \beta \in \mathbb{R}$ (note: the regularity is not supposed to be positive - and you get exactly the effect you are looking for in case of differentiability. The only problems appear if the regularities are integer: $\mathcal{C}^{1}$ is not exactly the space of continuously differentiable functions!).

Now we have a partition of the unity generated by $(\chi, \rho)$ through which we compute the norm of $f$. WLOG (up to a constant in the norm) we use a different partition for the norm of $g$: $(\chi, \psi)$ with $\psi$ supported in an annulus (larger than the one of $\rho$) such that $\psi \cdot \rho = \rho.$

Now we can start the computations: $$ ||f * g||_{\mathcal{B}^{\alpha + \beta}_{\infty , \infty}} = \sup_j 2^{(\alpha + \beta)j}||\Delta_j f * g||_{\infty}$$

We get $$\Delta_j f * g = \mathcal{F}^{-1}(\rho_j \mathcal{F}(f * g)) = \mathcal{F}^{-1}(\rho_j \mathcal{F}f \cdot \mathcal{F}g) =$$ $$ = \mathcal{F}^{-1}((\rho_j \mathcal{F}f) \cdot (\psi_j\mathcal{F}g)) = (\Delta_j f )*( \Delta_j g)$$

Hence we get that:

$$||f * g||_{\mathcal{B}^{\alpha + \beta}_{\infty , \infty}} = \le \sup_j 2^{(\alpha + \beta)j}||(\Delta_j f )*( \Delta_j g) ||_{\infty} \lesssim \sup_j 2^{(\alpha + \beta)j}||\Delta_j f ||_{\infty} \cdot || \Delta_j g||_{\infty} \le$$ $$\le ||f ||_{\mathcal{B}^{\alpha}_{\infty , \infty}} \cdot ||g ||_{\mathcal{B}^{\beta}_{\infty , \infty} }$$

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    $\begingroup$ Your proof looks satisfactory, but can you give some standard reference? $\endgroup$ Commented May 22, 2017 at 15:13
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    $\begingroup$ @BartoszWróblewski There are standard references such as "Fourier Analysis and nonlinear PDEs" by Bahouri et al. and there is also much to be found in "Theory of Function Spaces" by Triebel. For example it is a standard (but of course not trivial) result that the norms generated by different partitions of the unity are equivalent, which I have implicitly used in this. Nonetheless I could not find this result stated like this anywhere, I suppose because it´s just so simple, which is why I took the time to answer. $\endgroup$
    – Kore-N
    Commented May 26, 2017 at 8:14

While the proof given by Kore-N is very nice and easy to extend to more general scales of function spaces, I thought someone may find it useful to have a more direct argument. So here it is.

Write $\Delta_h f(x) = f(x + h) - f(x)$. Suppose that $f \in C^\alpha$ and $g \in C^\beta$, and one of them — say, $g$ — is compactly supported. Since $$ \Delta_h^2 (f * g)(x) = (\Delta_h f) * (\Delta_h g) ,$$ we have $$ |\Delta_h^2 (f * g)(x)| \leqslant \|\Delta_h f\|_\infty \|\Delta_h g\|_1 \leqslant C |h|^\alpha |h|^\beta ,$$ where $C$ depends on the Hölder constants of $f$ and $g$ and the size of the support of $g$.

It is now a well-known fact that the above condition implies that $f * g$ is in the Hölder–Zygmund space $\Lambda^{\alpha + \beta}$, which:

  • coincides with the space of Hölder continuous functions when $\alpha + \beta < 1$;

  • coincides with the space traditionally denoted by $C^{1,\alpha+\beta-1}$ if $1 < \alpha + \beta < 2$;

  • is strictly larger than the class of Lipschitz-continuous functions, but it is reasonably close when $\alpha + \beta = 1$.

I am not aware of any simple proof of this fact. A very nice and reasonably elementary argument is given by Stein in his Singular Integrals and Differentiability Properties of Functions book (see Proposition V.8 there).


  • More generally, we may assume that $p, q, r \in [1, \infty]$ are such that $\frac{1}{p} + \frac{1}{q} = 1 + \frac{1}{r}$ (as in Young's inequality), and $\|\Delta_h f\|_p \leqslant C |h|^\alpha$ and $\|\Delta_h g\|_q \leqslant C |h|^\beta$. Then $\|\Delta_h^2 (f * g)\|_r \leqslant C' |h|^{\alpha + \beta}$.

  • The above argument clearly carries over to arbitrary locally compact Abelian groups.

  • Yet another approach uses the fact that Hölder–Zygmund spaces are interpolation spaces between the usual classes $C^k$, and proceeds as in the standard proof of Young's inequality by real interpolation.

  • $\begingroup$ Just curious, but a result like this should admit generalizations. For example, if $f$ and $g$ lie in two regularity classes, then $f * g$ should lie in their "sum". I saw this reasoning being used, for example, in your paper with Prof. Grzywny. (Theorem 1.7, "Potential kernels, probabilities of hitting a ball, harmonic functions and the boundary Harnack inequality for unimodal Levy processes"). Are you aware of such generalizations, and is the current result "tight" in that we cannot (in general) expect $h \in C^{\alpha+\beta+\epsilon}$ for some $\epsilon>0$? $\endgroup$ Commented Aug 4, 2023 at 4:49
  • $\begingroup$ To be more precise on how the reasoning was used, the Poisson kernel (of a domain) is some form of a convolution between the jump kernel and the Green function of the domain, as mentioned in equation (2.3). Assuming regularity of the jump kernel/Levy measure (assumption of Theorem 1.7), we obtain regularity of the Poisson kernel and therefore of any harmonic function (which is a convolution of the Poisson kernel with the initial condition). $\endgroup$ Commented Aug 4, 2023 at 4:56
  • $\begingroup$ @SarveshRavichandranIyer: Regarding generalizations, I think Besov spaces are the answer. "Tightness" — I think $C^{\alpha+\beta+\epsilon}$ is already too much, but I do not have a counterexample at hand. Maybe Samko gives one in his book on hypersingular integrals? $\endgroup$ Commented Aug 4, 2023 at 9:14
  • $\begingroup$ Thank you for your response, I will look up Besov spaces and Samko's book. There should be something in this regard over there. $\endgroup$ Commented Aug 4, 2023 at 10:04

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