# Time averages and differentiability

Let $\varphi_t : M \rightarrow M$ be a smooth flow on a smooth manifold $M$. We may assume (although I'm not sure if this is important) that the flow preserves a smooth volume form on $M$. Given a continuous function $f : M \rightarrow \mathbb{R}$, is it true that if ALL the (partial) time averages $$A_T f(x) := {1 \over T} \int_0^T f(\varphi_t (x)) \, dt \hskip 1cm (T > 0)$$ are smooth functions, then $f$ is also smooth?

Remark. This is obviously true for the trivial action and it is non-trivially true for the case where $M = \mathbb{R}$ and the flow is the usual flow by translations. That's all I know for now.

Motivation. In the first volume of Dunford and Schwartz, where they motivate the Ergodic theorem (page 657), they state:

What is significant and measurable in the laboratory is not the quantity $f(\phi_t(x))$ but its average value $${1 \over T} \int_0^T f(\varphi_t (x)) \, dt$$ computed over a certain time interval $0 \leq t \leq T$.

I'm asking whether knowing the regularity of all those average values says something about the regularity of the function, or "observable" $f$. It would be interesting to consider $f$ to be just locally integrable and then the question ties in a (very) little with Wiener's differentiation theorem.

• sorry for posting it as an answer, but i don't have the right to comment. since you are not requiring anything for the limit $T \rightarrow \infty$, the normalization of the time integral by $T^{-1}$ does not really serve a purpose, does it? – jesus Jul 2 '15 at 13:14
• Not really. It only serves in pointing out that as T goes to zero, $A_T f(x) = f(x)$. The (very loose) analogy here is not with Birkhoff's ergodic theorem, but with Wiener's differentiation theorem. – alvarezpaiva Jul 3 '15 at 8:05
• I don't expect this to be true as averaging tends to improve the regularity of a function. Do you have a reference for the non-trivial result in your remark? – K Hughes Jul 3 '15 at 11:01
• The non-trivial result is equivalent to the statement that if $f$ is a continuous, real-valued function on the reals and $f(x+ T) - f(x)$ is smooth for every fixed $T$, then $f$ is smooth. I had asked this to my colleague, Jean-François Burnol and he gave a beautiful proof which, as he described, is an ode to Baire's theorem. He even proved it in the case $f$ is just assumed to be a distribution. I don't know if he published it or plans to publish it, but it is nicely written and you can ask him for a copy if you like. – alvarezpaiva Jul 3 '15 at 12:55
• I would be grateful to see a copy if possible. Thanks! – K Hughes Jul 10 '15 at 13:39

$$\textbf{Update with some corrections}$$:

Unfortunately, my original answer had a serious gap. Sorry for the mix-up. The argument I gave only works when we have estimates on $$A_T f(x)$$ that are uniform in $$T$$. However, all the hard work was not useless as it made clear the properties that a counterexample would have. Without further ado, here is an example for which $$A_T f(x)$$ is smooth while $$f(x)$$ is not.

Let $$M= \mathbb{R^2} =(t,x)$$ with the flow $$\phi_s (t, x) := (t+s, x)$$.

Let $$\Psi$$ be the following function

$$\Psi(x) = \begin{cases} 0 & x\leq 0 \\ e^{\frac{-1}{x^2}} & x > 0 \\ \end{cases}$$

In particular, we want $$\Psi$$ to be a smooth function which vanishes to all orders at $$0$$. Let $$f$$ be the following function:

$$f(t,x)= \begin{cases} e^{\frac{-t^2}{(\Psi(x))^2}} & x >0 \\ 0 & x \leq 0 \end{cases}$$

The function $$f$$ is not smooth. In fact, it is not even continuous at $$(0,0)$$. To show that $$A_T f(x)$$ is smooth for any $$T$$, we must only show that it depends smoothly on $$x$$ near $$(0,0)$$. For fixed $$T$$>0, we have the following:

$$A_T(f(t_1,x))= \frac{1}{T} \int_{t_1}^{t_1+T} f(t,x) dt < \frac{\sqrt{ \pi}}{T} \Psi(x) = o(x^k) ~ \forall k >0$$

From this, it follows that $$\frac{\partial^k A_T f}{\partial x^k}(t,0) \equiv 0$$, so the derivative of $$A_T f$$ exists to all orders at $$(0,0)$$ for $$T>0$$. The smoothness of $$A_T f(x)$$ elsewhere is immediate because $$f$$ is smooth away from the origin.

For an example where $$f(x)$$ is continuous but not smooth, we can consider the function $$\hat f(t,x) = x \cdot f(t,x)$$. Here, the same analysis holds except that $$f$$ is continuous but not differentiable at $$(0,0)$$.

$$\textbf{Original answer with the some corrections}$$

This is not a full answer, but I believe that your colleague's result does all the heavy lifting away from the fixed points of the flow, at least once you have good estimates on $$A_T f(x)$$. The case near a fixed point of a flow seems more difficult. I'm not sure how that works exactly, but if one can understand it in the $$1$$-dimensional case (i.e. when the flow has a limit point), this might give some good insight. I broke the answer into several sections to help make it more readable.

$$\textbf{Finding the right coordinates}$$

In this case, the smooth flow locally foliates your manifold, so around any point $$p$$ one can use the inverse function theorem to choose local coordinates $$\{ x_0, x_1, \ldots, x_{n-1} \}$$ for which the flow acts as $$\phi_t(x_0) = x_0+t$$ and $$\phi_t(x_i) = x_i$$ otherwise. By Professor Burnol's argument (which I'm black-boxing for this argument) $$f( x_0, x_1, \ldots, x_{n-1})$$ depends smoothly on $$x_0$$ when all the other coordinates are fixed.

What remains to show is that it depends smoothly on $$x_1, \ldots, x_n$$. We start with the case where $$x_0$$ is some fixed value, say 0. In this case, the $$A_T f(x)$$ is exactly the following:

$$\frac{1}{T}\int_0^T f(t, x_1, \ldots, x_n) ~dt$$

For the rest of this, I will use the parameter $$t$$ for $$x_0$$ and refer to the rest of the coordinates collectively as $$\mathbf{x}$$.

We know that $$A_T f(x)$$ depends smoothly on $$\mathbf{x}$$ and further that $$f$$ depends smoothly on $$t$$. What remains to show is that $$f$$ depends smoothly on $$\mathbf{x}$$. We can't take derivatives with respect to $$\mathbf{x}$$, so we need to estimate some difference quotients. I will be using the Euclidean norm on the $$\mathbf{x}$$ coordinates for my estimates.

$$\textbf{The uniform Lipschitz estimate}$$

Take two points $$\mathbf{x}^1$$ and $$\mathbf{x}^2$$. Since $$A_T f(x)$$ is smooth, it is locally Lipschitz, and so we have the following estimate:

$$\frac{1}{T \| \mathbf{x}^1- \mathbf{x}^2 \| } | \int_0^T f(t, \mathbf{x}^1) - f(t, \mathbf{x}^2) ~dt | < \mathcal{C}$$

In order for the rest of the argument to work, we must assume that this estimate is uniform in the choice of $$T$$, $$\mathbf{x}^1$$ and $$\mathbf{x}^2$$. This allows us to pick $$T>0$$ small enough so that that for our fixed $$\mathbf{x}^1$$ and $$\mathbf{x}^2$$, we have the following:

$$\| f(t, \mathbf{x}^i) -f(s, \mathbf{x}^i) \|_{C^2} < \epsilon$$ for $$0.

The choice of $$T$$ may depend on $$\mathbf{x}^1$$ and $$\mathbf{x}^2$$. In particular, we don't expect to be able to do this uniformly in $$\mathbf{x}$$, but that's okay because we are only doing it for two points.

Then, we have the following estimate.

$$\begin{eqnarray} | \frac{1}{T} \int_0^T \frac{ f(t, \mathbf{x}^1) - f(t, \mathbf{x}^2)}{\| \mathbf{x}^1- \mathbf{x}^2 \| } ~dt | & \geq & | \frac{1}{T} \frac{ T(f(0,\mathbf{x}^1) - f(0,\mathbf{x}^2))}{\| \mathbf{x}^1- \mathbf{x}^2 \|} | \\ & & - \frac{1}{T} \left( \frac{\partial f}{\partial t} (0, \mathbf{x}^1)+ \frac{\partial f}{\partial t} (0, \mathbf{x}^2) + 2\epsilon \right) \frac{T^2}{2} \\ & \geq & | \frac{f(0,\mathbf{x}^1) - f(0,\mathbf{x}^2)}{\| \mathbf{x}^1- \mathbf{x}^2 \|}| - \left( \frac{\partial f}{\partial t} (0, \mathbf{x}^1)+ \frac{\partial f}{\partial t} (0, \mathbf{x}^2) + 2\epsilon \right) T \\ \end{eqnarray}$$

Letting $$T$$ be go to zero using the uniformity of the earlier estimate while still fixing $$\mathbf{x}^1$$ and $$\mathbf{x}^2$$ for now, our work shows the following.

$$| \frac{f(0,\mathbf{x}^1) - f(0,\mathbf{x}^2)}{\| \mathbf{x}^1- \mathbf{x}^2 \|}| \leq \mathcal{C}$$

Notice that this $$\mathcal{C}$$ is the same constant as before in the estimate of $$A_T f(x)$$. Repeating this for arbitrary pairs of $$\mathbf{x}$$, this gives a uniform Lipschitz estimate on $$f$$ (at $$t=0)$$ in terms of the $$\mathbf{x}$$ coordinates.

$$\textbf{A sketch of how to use induction for higher order regularity}$$

In fact, we can repeat essentially the same argument but with three points $$\mathbf{x}^1$$, $$\mathbf{x}^2$$ and $$\lambda \mathbf{x}^1 + (1- \lambda) \mathbf{x}^2$$ where instead of using the difference quotient, we use the second order difference quotient. This should yield a uniform $$C^{1,1}$$ estimate on $$f$$ in terms of the smoothness of $$A_T f(x)$$. With a uniform $$C^{1,1}$$ estimate, this shows that the function $$f$$ was actually differentiable in terms of $$\mathbf{x}$$ all along. You can induct on all orders to get smoothness of $$f$$ in $$\mathbf{x}$$. This shows that $$f$$ is smooth in both $$t$$ and $$\mathbf{x}$$.