# If $f$ be Riemann integrable on $[0,x]\,\forall\,x>0$ then $\liminf\limits_{x\to\infty}f(x)\leq\liminf\limits_{x\to\infty}\frac{1}{x}\int_{0}^{x}f$

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.

• The first step “it follows that...” is wrong. Take $z=f=y=1$. – 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 $$T \to \infty$$).
• 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. – John Omielan Jun 19 '19 at 21:20
• @JohnOmielan Thanks for the comment. There was a typo and the denominator was supposed to be $x$, not $T$. – Kavi Rama Murthy Jun 19 '19 at 23:09
• 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 $\lim_{T \to \infty}g(T)$. – RRL Jun 20 '19 at 1:34
• @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 $T \to \infty$. – Kavi Rama Murthy Jun 20 '19 at 5:13