Suppose $f:\mathbb{R} \to \mathbb{R}$ has the property that for every fixed $t\in\mathbb{R}$ the function $$ g_t : x \mapsto f(x) - f(x-t) $$ is $C^\infty(\mathbb{R})$. Does it follow that $f$ is smooth?

Edit: The answer is no in generality (see answer below), but what if we impose the condition that $f$ is measurable?

You can translate this as a question in group cohomology: The map $t \mapsto g_t$ is a $1$-cocycle (not assumed continuous) for $\mathbb{R}$ with values in the $\mathbb{R}$-module $C^\infty(\mathbb{R})$ (with translation). Is the cohomology group $H^1(\mathbb{R}; C^\infty(\mathbb{R}))$ trivial?

(The question is motivated by a technical point relating to distributions on groupoids, which I won't go into here.)

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    $\begingroup$ The function $f$ is smooth everywhere if it is smooth near one point. To prove smoothness at $z$, choose $t=z-x_0$ where $x_0$ is the point of smoothness. $\endgroup$ Oct 9, 2014 at 8:55

3 Answers 3


At least under the axiom of choice the answer is negative: The are discontinuous additive functions $f;\mathbb R \to \mathbb R$ (as far as I remember, that's why Hamel inveted Hamel basis) and for such a function $g_t(x)=f(x)-f(x-t)$ is constant for all $t$.

  • $\begingroup$ Excellent point. I'd like to add conditions to eliminate Axiom of Choice-type constructions. $\endgroup$ Oct 10, 2014 at 15:16

Hamel basis is "not of this world":-)

It becomes more interesting if we make some mild a priori assumption, like $f\in L^1(R)$, and $g_t^{(n)}\in L^1(R)$. Then the answer is "yes", $f$ must be smooth.

We apply the Fourier transform. We obtain $G_t(s)=(1-e^{its})F(s)$. As $g$ is $C^\infty$, we have that $|G_t(s)|\leq K_n|s|^{-n}$, for every $n$.

Let us prove that $|F|=O(|s|^{-n})$ for an arbitrary fixed $n$. Let $h(s)=s^nF(s)$. Suppose by contradiction that there is a sequence $s_k\to\infty$ such that $h(s_k)\to\infty$.

As we have the inequality $|h(s)|\leq |C_t/\sin(st)|$ for ALL $t$, the sequence $s_k$ must have a strange property: it must be close to ALL arithmetic progressions.

Does such a sequence exist? Precise formulation: does these exist a sequence $s_k\to\infty$ such that for every positive $a$ the sequence $S_k=\min_m|s_k-am|$ tends to $0$ ?

No it does not. Suppose it exists, and consider the integrals $$I_k=\int_1^2 \min_m|s_k-mx|dx.$$ The function under the integral sign is bounded (does not exceed $1$). If our sequence has the desired property, then $I_k\to 0$ (dominated convergence). But this is a contradiction because the integrals evidently $\geq 1$.

The argument in the previous paragraph is due to user fedja.

It follows from this proof that positive measure set of $t$ is enough, while any countable set of $t$ is not enough.

EDIT. It is funny what happens if we require bounded support of $f$, in addition to $L^1$. Then $F$ is an entire function of exponential type, bounded on the real line, so by Bernstein's theorem, $F'$ is bounded and we obtain that $f$ is smooth if $g_t$ is smooth for only ONE value of $t$.


Conjecture: if $f$ is locally $L^1$ and there is a set $E$ of positive measure such that $g_t$ are smooth for $t\in E$ then $f$ is smooth. Same is probably true with $E$ of second Baire category.

  • $\begingroup$ Don't you need further growth conditions for the derivatives of $g_t$ in order to get the bound for $|G_t(s)|$? $\endgroup$ Oct 10, 2014 at 6:27
  • $\begingroup$ @Jochen Wengenroth: you are right. We need derivatives in $L^1$. I made a correction. $\endgroup$ Oct 10, 2014 at 13:29
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    $\begingroup$ Nice answer. But do we need as strong as $L^1$ (which forces some kind of decay at infinity, at least in a measurable sense). What if $f$ is just measurable, for instance? $\endgroup$ Oct 10, 2014 at 15:33
  • $\begingroup$ @Bob Yuncken: Bob, there is a lot of possibilities for generalization. For example, $f$ can be periodic and locally $L^1$, or in Schwartz temperate distribution space. Some growth conditions are however necessary for Fourier analysis to work. It is actually strange that the answer to such question may depend on the growth of $f$ rather than just local properties. But I do not see how to get rid of the growth conditions completely. $\endgroup$ Oct 10, 2014 at 19:21
  • $\begingroup$ Yes, I though about the possibility of having $f$ as a tempered distribution, but it seems you still need that the $g_t$ are $L^1$ for the argument to work, which is rather restrictive. At the very least, you need is that $G_t$ be a function, and I'm not sure what condition on a slowly growing (or even bounded) smooth function is necessary for that. On the other hand, your argument seems to work fine for the analogous question on the circle. So if we can show that the problem is "local" then your solution solves everything. $\endgroup$ Oct 12, 2014 at 18:02

Here's a solution for $f\in L^1_{\mathrm{loc}}$, or even for $f$ a distribution, which avoids Fourier analysis by mollifying.

Let $\phi \in C^\infty_c(\mathbf{R})$ be a smooth bump function of total mass $1$. Define $\tilde{f} := \phi * f$. Then $f\in C^\infty(\mathbb{R})$. Moreover, $$ (f-\tilde{f})(x) = \int (f(x) - f(x-t))\phi(t)dt = \int g_t(x)dt. $$ So we're done if we can prove that that last integral is convergent in $C^\infty(\mathbb{R})$. Note that a priori it is only convergent as a distribution.

Here's one way to take care of that last logical point. A Baire category argument shows that for any $k\in\mathbb{N}$, there is some non-empty interval $[a,b]$ such that the $C^k$-norms of $\{g_t\}_{t\in[a,b]}$ are bounded. By translation, this is true for any translate of $[a,b]$ as well, and thus for any compact interval. Arzela-Ascoli, implies that $\{g_t\}_{t\in[a,b]}$ is compact in $C^{k-1}(\mathbb{R})$. From this and the fact that $t\mapsto g_t$ is weakly smooth, we can deduce that it is continuous in the $C^{k-1}$-topology, for all $k$.


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