Asymptotic behavior of integral MVT intermediate point

Question

If a function $f:[0,\infty) \to \mathbb{R}$ is continuous, the integral mean value theorem guarantees the existence of at least one point $\xi_x \in (0,x)$ such that

$$\int_0^x f(t) \, dt = f(\xi_x)x.$$

Observe some simple cases (with unique intermediate points). If $f(x) = x$ then $\xi_x = x/2$ and $\xi_x/x \to 1/2$ as $x \to 0$. If $f(x) = x^2$ then $\xi_x = x/ \sqrt{3}$ and $\xi_x/x \to 1/\sqrt{3}$ as $x \to 0.$

Now, consider

$$f(x) = \begin{cases} x^2 \sin(1/x), \,\,x >0 \\ 0, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0\end{cases}$$

and any $\xi_x \in (0,x)$ such that

$$\int_0^x f(t) \, dt = \xi_x^2 \sin(1/ \xi_x) x.$$

What can be said in this case about $\lim_{x \to 0} \frac{\xi_x}{x}?$

Addendum:

Using Taylor's theorem, I could show that if any $f$ is differentiable at $x = 0$ with $f'(0) \neq 0$, then $\xi_x / x \to 1/2.$ If any $f$ is twice differentiable at $x = 0$ with $f'(0) = 0$ and $f''(0) \neq 0$, then $\xi_x / x \to 1/\sqrt{3}.$ (This result generalizes if $f$ is $k$ times differentiable and the first $k-1$ derivatives vanish.)

The problem here is that $f'(0) = 0$ and $f''(0)$ does not exist.

Answer 1

For $x>0,$ define $F(x) = \int_0^x f(t)\, dt$ where $f(t) = t^2\sin (1/t).$ For each such $x,$ consider the set $$C(x) = \{c\in (0,x): F(x)/x=f(c)\}.$$ Here's my basic claim: All hell breaks loose. More precisely, given any $y\in [0,1]$ there are sequences $x_n \to 0^+$ and $c_n \in C(x_n)$ such that $c_n/x_n \to y.$ Another way to put it: your quotients $c/x,$ with $c\in C(x),$ are asymptotically dense in $[0,1]$ as $x\to 0^+.$

On the other hand, if we define $c^*(x) = \max \{c\in C(x)\},$ then $c^*(x)/x\to 1.$

For a proof, let's look at the case $F(x) =0,$ which is the same as saying $F(x)/x = 0.$ For which $x$ can this happen? Let $a_n = 1/(n\pi).$ Then an alternating series argument shows $F(a_{n})>0$ for $n$ even, $F(a_{n})<0$ for $n$ odd. By the intermediate value theorem, there exists $x_n\in (a_{n+1},a_{n})$ such that $F(x_n) = 0.$ Because $F'\ne 0$ on any $(a_{n+1},a_{n}),$ $x_n$ is the unique zero of $F$ in this interval.

What does $C(x_n)$ look like? This is easy: it's just the set of points $c$ to the left of $x_n$ where $f(c) = 0.$ That is precisely the set $a_k, k\ge n+1.$ It's now easy to see $c^*(x_n)/x_n \to 1$ as claimed. But we can say more: The sequence $a_k/x_n, k\ge n,$ which forms an infinite partition of $(0,x_n),$ has mesh size no more that $a_n-a_{n+1}\to 0.$ That is enough to give the "asymptotically dense" claim.

I've left out details of some of these sub claims, so feel free to ask about them.

Answer 2

Suppose there exists a differentiable function $g(x)=\xi_x$ (I think such $g(x)$ exists for large enough $x$ by inverse function theorem) so we have $\int_0^x f(x) \, dx = f\circ g(x) x...(0)$. Since $\lim_{x\to \infty} f(x)=\infty$, we should have $\lim_{x\to \infty}g(x)=\infty$. By L Hopital's rule, we have $$\lim_{x\to \infty}\frac{g(x)}{x}=\lim_{x\to \infty}g'(x)...(1)$$ Taking $\frac{d}{dx}$ on both sides of $(0)$, we have: $$f(x)=f\circ g(x)+xf'(g(x))g'(x)$$ OR $$\frac{f(x)}{x}=\frac{f\circ g(x)}{x}+f'(g(x))g'(x)$$ so $$g'(x)=\frac{\frac{f(x)}{x}-\frac{f\circ g(x)}{g(x)}\frac{g(x)}{x}}{f'(g(x))}$$ $\lim_{x\to \infty}\frac{f(x)}{x}=1$ and $\lim_{x\to \infty}f'(x)=1$ so $\lim_{x\to \infty}g'(x)=0$.

If we know $\lim_{x\to \infty}g'(x)=l$ exists, then $l=\frac{1}{2}$.

I think I still need to show why it exists...