Consider a real function $f: \mathrm{R} \to \mathrm{R}$, $f$ is piecewise right continuous, and this pair of equations:
$$ \sup_{x> a} \frac 1 {x-a} \int_a^x f(s) \, ds = \phantom{-}\infty \tag1\\ \inf_{x> a} \frac 1 {x-a} \int_a^x f(s) \, ds = -\infty. $$
A remark I found states: If (1) holds for some $a \in \mathrm{R}$, then it holds for all $a \in \mathrm{R}$.
I need help proving this remark.
I see two possible ways to tackle this problem:
- using the (first) mean value theorem for definite integrals
- using the fundamental theorem of calculus
I am, however, in both cases not sure how to deal with the required closed intervals and the open interval $(x, \infty)$ from the supremum/infimum.
Hopefully, someone can give me a hint.
Note first that $$ \sup_{x> a} \frac 1 {x-a} \int_a^x f(s) \, ds = +\infty $$ if and only if there exists a sequence $x_k\nearrow+\infty$ such that $$ \lim_{k\to\infty}\frac 1 {x_k-a} \int_a^{x_k} f(s) \, ds = +\infty. $$ Now take any $b\in\mathbb R$. Then $$ \begin{split} \lim_{k\to\infty}\frac 1 {x_k-b} \int_b^{x_k} f(s) \, ds &= \lim_{k\to\infty}\frac {x_k-a} {x_k-b} \cdot\frac 1 {x_k-a} \left(\int_a^{x_k} f(s) \, ds -\int_a^b f(s) \, ds\right)\\ &=\lim_{k\to\infty}\frac 1 {x_k-a} \int_a^{x_k} f(s) \, ds= +\infty \end{split} $$ since $$ \lim_{k\to\infty}\frac {x_k-a} {x_k-b}=1 $$ and $$ \lim_{k\to\infty}\frac 1 {x_k-a} \int_a^b f(s) \, ds=0. $$
You can do the same for the infimum.