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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.

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

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  • $\begingroup$ 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 at 21:20
  • $\begingroup$ @JohnOmielan Thanks for the comment. There was a typo and the denominator was supposed to be $x$, not $T$. $\endgroup$ – Kabo Murphy Jun 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 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 $T \to \infty$. $\endgroup$ – Kabo Murphy Jun 20 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 at 5:19

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