Let $f$ be Riemann integrable on $[0,x]$ for all $x>0$. Prove that $$\liminf\limits_{x\to\infty}f(x)\leq\liminf\limits_{x\to\infty}\frac{1}{x}\int_{0}^{x}f.$$

From the definition of $\liminf$, I need to prove that $$\lim\limits_{y\to\infty}\inf\limits_{x\geq y}f(x)\leq\lim\limits_{y\to\infty}\inf\limits_{x\geq y}\frac{1}{x}\int_{0}^{x}f.$$ Fix $y>0$. Then for any fixed $z\geq y$ it follows that $$\inf\limits_{x\in[y,z]}f(x)\leq\frac{1}{z}\int_{0}^{z}f.$$ Since this is true for all $z\geq y$, we have that $$\inf\limits_{z\geq y}\inf\limits_{x\in[y,z]}f(x)\leq \inf\limits_{z\geq y}\frac{1}{z}\int_{0}^{z}f.$$

Since the above holds for all $x>0$, it follows that $$\lim\limits_{y\to\infty}\inf\limits_{x\geq y}f(x)\leq\lim\limits_{y\to\infty}\inf\limits_{x\geq y}\frac{1}{x}\int_{0}^{x}f.$$

I am almost certain that I made a mistake handling all the infima, and that I failed to prove the result.

Could anybody please point out if I made a mistake. And if I did, could anyone give me a hint on how to look at the problem. Please don't solve it for me.

  • $\begingroup$ The first step “it follows that...” is wrong. Take $z=f=y=1$. $\endgroup$ – Alex R. Jun 18 '19 at 4:31

For $x>T>0$ we have $\frac 1 x \int_0^{x}f(t)dt =\frac 1 x \int_0^{T}f(t)dt+\frac 1 x \int_T^{x}f(t)dt \geq \frac 1 x \int_0^{T}f(t)dt+ g(T) \frac {x-T} x$ where $g(T)$ is the infimum of $f(t)$ for $t \geq T$. Now just take limit on both sides as $x \to \infty$. The first term tends to $0$ and the second term tends to $g(T)$ (which tends to $\lim \inf_{x \to \infty} f(x)$ as $T \to \infty$).

  • $\begingroup$ In the last part of your inequalities, note $\frac{1}{x}\int_{0}^{T}f(t)dt \ge g(T)\frac{x-T}{x}$, with no other more stringent condition possible which I'm aware of with the general conditions provided here. Thus, assuming $g(T) \gt 0$, since $x \gt T \gt 0$, then $g(T)\frac{x-T}{x} \lt g(T)\frac{x-T}{T}$, so you can't do the replacement you did. However, it's valid if $g(T) \le 0$, but I don't see how you may assume this here. Please let me know if I'm missing or misunderstanding something. Thanks. $\endgroup$ – John Omielan Jun 19 '19 at 21:20
  • $\begingroup$ @JohnOmielan Thanks for the comment. There was a typo and the denominator was supposed to be $x$, not $T$. $\endgroup$ – Kavi Rama Murthy Jun 19 '19 at 23:09
  • $\begingroup$ What also is not clear is the statement that $g(T) \frac{x-T}{x}$ tends to $\liminf_{x \to \infty} f(x)$ as $x \to \infty$. Certainly not with fixed $T< x$ where the limit is $g(T)$. I suppose you can say let $T = x \epsilon$ in which case the limit as $x \to \infty$ is $\liminf_{x \to \infty} f(x) (1-\epsilon)$ and the LHS exceeds this for any $\epsilon > 0$. Or $\liminf$ of the LHS exceeds $g(T)$ for any $T$ and thus exceeds $\lim_{T \to \infty}g(T)$. $\endgroup$ – RRL Jun 20 '19 at 1:34
  • $\begingroup$ @RRL If you let $x \to \infty$ (with $T$ fixed) you get $\lim \inf \frac 1 x\int_0^{x}f(t)dt \geq g(T)$ for every $T$. Now let $T \to \infty$. $\endgroup$ – Kavi Rama Murthy Jun 20 '19 at 5:13
  • $\begingroup$ @KaviRamaMurthy: Sure -- that was what I said in the last sentence of my comment. Good answer -- just a little unclear at the end. (+1) $\endgroup$ – RRL Jun 20 '19 at 5:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.