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Christian Remling
Christian Remling
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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 $<1/2$).

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.

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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Christian Remling
Christian Remling
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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 $$ \lim_{t\to 0+}\frac{1}{t}\int_0^1d\mu(s) \int^s_{s-t} dx\, f(x) =\int_0^1 f(s)\, d\mu(s) , $$$$ \lim_{t\to 0+}\frac{1}{t}\int_0^1d\mu(s) \int^s_{s-t} dx\, f(x) =\int_0^1 f(s)\, d\mu(s)=\frac{1}{2} , $$ 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.

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 $$ \lim_{t\to 0+}\frac{1}{t}\int_0^1d\mu(s) \int^s_{s-t} dx\, f(x) =\int_0^1 f(s)\, d\mu(s) , $$ 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.

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.

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Christian Remling
Christian Remling
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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 and bounded, $(1/t)\int_{s-t}^s f(x)\, dx$ increases to a limit as $t\to 0+$ (at Lebesgue almost every $s$, the limit equals $$ f(s-t)\le \frac{1}{t}\int_{s-t}^s f(x)\, dx\le f(s) , $$ and thus $f(s)$, by Lebesgue's differentiation theorem, but this doesn't matter here)$\lim_{t\to 0+}(1/t)\int_{s-t}^s f(x)\, dx= f(s)$. Hence monotone convergence shows that $$ \int_0^1d\mu(s)\frac{1}{t} \int^s_{s-t} dx\, f(x) $$$$ \lim_{t\to 0+}\frac{1}{t}\int_0^1d\mu(s) \int^s_{s-t} dx\, f(x) =\int_0^1 f(s)\, d\mu(s) , $$ converges. Finally,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.

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 and bounded, $(1/t)\int_{s-t}^s f(x)\, dx$ increases to a limit as $t\to 0+$ (at Lebesgue almost every $s$, the limit equals $f(s)$, by Lebesgue's differentiation theorem, but this doesn't matter here). Hence monotone convergence shows that $$ \int_0^1d\mu(s)\frac{1}{t} \int^s_{s-t} dx\, f(x) $$ converges. Finally, $(1/t)\int_1^{1+t} \ldots \to 0$ since $\mu(1,1+t)\to 0$ and $f$ is bounded.

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 $$ \lim_{t\to 0+}\frac{1}{t}\int_0^1d\mu(s) \int^s_{s-t} dx\, f(x) =\int_0^1 f(s)\, d\mu(s) , $$ 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.

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Christian Remling
Christian Remling
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Christian Remling
Christian Remling
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