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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 $f$?

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    $\begingroup$ Your main estimate is not uniform in $t$ (as you claim). In fact, the smaller $t>0$ is, the larger becomes the right-hand side (going to $\infty$ as $t\to0$). For this reason, it is not possible to conclude from this estimate that you can exchange the limits. Roughly speaking,your function $a$ must depend on $t$ to get a good estimate for $\lvert g(t)-T(a)(t)\rvert$ (because you have no uniform estimate), but if $a$ depends on $t$, you cannot estimate $\lvert T(a)(t)-h(a)\rvert$. In fact, my intuition is that the whole approach with Weierstraß does not work. $\endgroup$ Commented Jan 6, 2024 at 13:23
  • $\begingroup$ The claim is that if $f$ is continuous and of bounded variation, then the limit exists and is what one would expect, namely $\frac 12 (f(1)^2-f(0)^2)$. The reason is that $f$ can be regarded as a distribution and so the limit of $\frac 1t(f(x+t)-f(t))$ exists and is $f’$ (derivative and convergence in the distributional sense). Since one can always pull such limits under the integration sign when such parametrised integrals exist, it remains only to show that the product $f(x)f’(x)$ exist as a distribution and is integrable. This is true under the above assumptions. $\endgroup$ Commented Jan 7, 2024 at 12:13
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    $\begingroup$ You talk about distributions, but the OP only wants to work with Riemann, not Lebesgue integrals. This complicates things. Furthermore, one needs a supplementary argument to show that the product $f f'$ exists when $f$ is not smooth and $f'$ is understood as a distribution: the product between smooth functions and distributions is well defined, but the product between continuous functions with bounded variation and distributions is not. One should probably first show that the distribution $f'$ is, in fact, a Borel measure; only then would the product $f f'$ make sense. Too complicated! $\endgroup$ Commented Jan 7, 2024 at 14:48
  • $\begingroup$ It was explained in an introductory remark that this wasn't an answer but just a comment to add the info that monotonicity isn't necessary and that there was an alternative approach which might be of interest to the OP.. This was deleted by a moderstor, thus mutllating the modest contribution which had been made in good faith $\endgroup$ Commented Jan 8, 2024 at 4:44

3 Answers 3

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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 $<1/2$).

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    $\begingroup$ If Lebesgue integrals are off limits (as indicated in the OP), this should also be fine when the integrals $\int d\mu\ldots$ are interpreted as Riemann-Stieltjes integrals (though this seems about as meaningful as being asked to play a piano piece, touching the keys only with one's nose). $\endgroup$ Commented Jan 6, 2024 at 16:44
  • $\begingroup$ Nice argument, but the problem does not seem to be solved completely: For absolutely continuous $f$ with $f(0)=0$ and $f(1)=1$, it seems to be rather straightforward to see that the latter integral is $1/2$. Can this be shown for the general case as well? (In fact, it seems plausible for the devli's staircae, but I do not see a rigorous proof.) $\endgroup$ Commented Jan 6, 2024 at 21:23
  • $\begingroup$ @MartinVäth: This is immediate from a sufficiently general version of the substitution rule since $\mu$ is the image measure of Lebesgue measure under $f:[0,1]\to[0,1]$, so $\int_0^1 f\, d\mu = \int_0^1 x\, dx$. See here: en.wikipedia.org/wiki/… $\endgroup$ Commented Jan 6, 2024 at 21:29
  • $\begingroup$ @MartinVäth: Or even integration by parts: $\int_0^1 f\, df = f^2\bigr|_0^1-\int_0^1 f\, df$: en.wikipedia.org/wiki/… $\endgroup$ Commented Jan 6, 2024 at 21:31
  • $\begingroup$ Thanks: So far, I was only aware that integration by parts and substitution rule hold for absolutely continuous (substitution) functions. Thanks for the link to the general case. $\endgroup$ Commented Jan 6, 2024 at 21:34
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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 $\epsilon \to 0$.

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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 $f^2$.

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