Using the Stone-Weierstrass theorem to solve an integral limit

Question

The following question was posted on math stack exchange here but it got no answers

Let $c\in (1, +\infty)$ and $f \colon [0, c] \to \mathbb{R}$ be a continuous and monotonically increasing function with $f(0)=0$ and $f(1)=1$. Calculate the following limit $$\lim_{t \to 0^+} \frac{1}{t}\int_0^1 f(x)(f(x+t)-f(x))dx$$ (the integral is assumed to be a Riemann integral, it is supposed to be solved without Lebesgue integrals)

My approach to solving the problem:

The problem is trivial if we assume that $f$ is differentiable. If we do, then by the mean value theorem, there exists, for each $x \in [0, 1]$ a $c_t \in [x, x+t]$, such that $f(x+t)-f(x)=f'(c_t)t$. Since $f$ is continuous and $f'$ is bounded, so is $f(x)f'(c_t)$, for all $t$ near $0$. So our limit becomes $\lim_{t \to 0^+} \int_0^1 f(x)f(c_t)dx$. By Arzela's theorem for uniformly bounded integrals, we can interchange the limit and the integral and we get $\int_0^1 f(x)f'(x)dx=0.5(f(1)^2-f(0)^2)$.

Since every continuous function can be uniformly approximated by polynomials(Stone-Weierstrass theorem), then the set of differentiable functions with bounded derivative are dense in the set of continuous functions with the $||\cdot||_{\infty}$ norm, so for any continuous function $f_0$, we can find differentiable functions arbitrarily "close" to it.

These observations lead to considering the following function. Let $C$ be the Banach space of continuous functions with the $||\cdot||_{\infty}$ norm. Let $T \colon C \times (0, +\infty) \to \mathbb{R}, T(f)(t) = \frac{1}{t} \int_0^1 f(x)(f(x+t)-f(x))dx$. Consider now, $u \in C$ and $v \in C$ fixed and $t$ an arbitrary positive real number. Then $$ |T(u)(t)-T(v)(t)| \leq \frac{1}{t} \int_0^1 | u(x)u(x+t)-u(x)v(x+t)+u(x)v(x+t)-v(x)v(x+t)+v(x)^2-u(x)^2 |dx \leq \frac{1}{t}\left(\int_0^1|u(x)||u(x+t)-v(x+t)|dx+\int_0^1|u(x)-v(x)||v(x+t)|dx + \int_0^1|u(x)-v(x)||u(x)+v(x)|dx\right) \\ \leq \frac{1}{t} \left(\|u\|_{\infty}\|u-v\|_{\infty}+\|u-v\|_{\infty}\|v\|_{\infty} + \|u-v\|_{\infty}\|u+v\|_{\infty}\right) \leq \frac{1}{t} \left( \|u-v\|_{\infty} \left( \|u\|_{\infty}+\|v\|_{\infty} + \|u+v\|_{\infty} \right) \right) \leq \frac{1}{t} \left( 2\|u-v\|_{\infty} \left( \|u-v\|_{\infty}+2\|v\|_{\infty}\right) \right) $$

Therefore $T$ is continuous in $f$ uniformly in $t$. Because of $T(f)(t)$ converges uniformly to some function $g(t)$. Therefore, $$ \lim_{t \to 0} \lim_{f \to f_0} T(f)(t) =\lim_{f \to f_0} \lim_{t \to 0} T(f)(t) $$

To justify the interchange of limits, let's denote $g(t)\colon=\lim_{f \to f_0} T(f)(t)$, $h(f)= \lim_{t \to 0} T(f)(t)$ and $l=\lim_{f \to f_0} \lim_{t \to 0} T(f)(t)$. Now let $a$ be a function such that $\|a-f_0\|_{\infty}<\delta$. We can choose such $a$ because $f_0$ is an accumulation point. We can see that $$ |g(t)-l| \leq |g(t)-T(a)(t)+T(a)(t)-h(a)+h(a)-l| \leq |g(t)-T(a)(t)| + |T(a)(t)-h(a)| + |h(a)-l| $$ And each term can be made arbitrarily small.

Since the differentiable functions are dense in $C$, we can assume WLOG that $f$ is differentiable in the limits above. The question in the problem is equivalent to $\lim_{t \to 0} \lim_{f \to f_0} T(f)(t)$. And by interchanging limits, the result is trivial.

Is my proof correct? To me it seems like it is, but I never used the monotonicity of $f$?

Answer 1

Since $f$ is increasing, $\mu(0,x)=f(x)$ defines a (here: continuous) measure on $[0,c]$, which we can also view as a measure on $\mathbb R$ with support in this set. By Fubini, the RHS equals $$ \frac{1}{t}\int_0^1 dx\, f(x)\int_x^{x+t} d\mu(s)=\frac{1}{t}\int_0^{1+t}d\mu(s)\int^s_{s-t} dx\, f(x) . $$ Since $f$ is increasing, $$ f(s-t)\le \frac{1}{t}\int_{s-t}^s f(x)\, dx\le f(s) , $$ and thus $\lim_{t\to 0+}(1/t)\int_{s-t}^s f(x)\, dx= f(s)$. Hence monotone convergence shows that $$ \lim_{t\to 0+}\frac{1}{t}\int_0^1d\mu(s) \int^s_{s-t} dx\, f(x) =\int_0^1 f(s)\, d\mu(s)=\frac{1}{2} , $$ and $(1/t)\int_1^{1+t} \ldots \to 0$ since $\mu(1,1+t)\to 0$ and $f$ is bounded, so the original expression has the same limit.

Added later: We don't really need $f$ to be continuous here. The same calculation, in a slightly more careful version, also shows that in general $$ \lim_{t\to 0+} \frac{1}{t}\int_0^1 f(x)(f(x+t)-f(x))\, dt = \int_{[0,1]}f(s-)\, d\mu(s) , $$ where now $\mu([0,x])=f(x+)$ (and the limit could now be $<1/2$).

Answer 2

Consider first $g$ continuous, strictly increasing in $[0,1]$ with $g(0)=0$, $g(1)=a$ $$ \frac {1}{t}\int_0^1 f(x)(g(x+t)-g(x))\, dx=\frac{1}{t} \int_0^1 f(x) dx \int_{g(x)}^{g(x+t)} ds=\frac{1}{t}\int_0^{g(1+t)} ds \int_{g^{-1}(s)-t}^{g^{-1}(s)} f(x) dx $$ which tends to $\int_0^a f(g^{-1}(s)) ds$ as $t \to 0$ (one has to extend also $f$ outside $[0,1]$). If $f=g$ we get $\int_0^a s ds= \frac{a^2}{2}$. If is not stricly monotonic, apply this to $f(x)+\epsilon x$ and let $\epsilon \to 0$.

Answer 3

I understand the aim is an elementary proof; here is one.

Let $\omega$ be a modulus of continuity for $f$ on $[0,c]$. Then, for all $0<t\le c-1$

$$\int_0^1|f(x+t)-f(x)|^2dx\le \omega(t)\int_0^1\big(f(x+t)-f(x)\big)dx =$$$$= \omega(t)\bigg[ \int_1^{1+t} f(x)dx-\int_0^tf(x) dx\bigg]\le \omega(t)t\big(f(1+c)-f(0)\big)=o(t),$$ as $t\to 0$. So
$$\int_0^1 f(x)\big(f(x+t)-f(x)\big)dx=\int_0^1 f(x+t)\big(f(x+t)-f(x)\big)dx+o(t)=$$$$=\frac12 \bigg[\int_0^1 f(x)\big(f(x+t)-f(x)\big)dx+\int_0^1 f(x+t)\big(f(x+t)-f(x)\big)dx\bigg]+o(t) =$$ $$=\frac12 \int_0^1 \big(f(x+t)^2-f(x)^2\big)dx +o(t)= \frac12\bigg[\int_1^{1+t} f(x)^2dx-\int_0^tf(x)^2 dx\bigg] +o(t)=$$ $$=\frac12\big( f(1)^2 -f(0)^2\big)t +o(t),$$ by the Fundamental theorem of Calculus for $f^2$.