Let $s\_1, s\_2, \dotsc$ be a real sequence and define
$$\sigma\_n = \frac{s\_1 + s\_2 + \dotsb + s\_n}{n}.$$ The inequality
$$\operatorname{lim sup}\sigma\_n \leq \operatorname{lim sup} s\_n$$
is well known and trivially proved.


Consider a real valued continuous function $f(x)$ defined on the positive real line and oscillating in sign infinitely often.
The integral analog of the Cesàro mean for a sequence is the Cesàro mean of the integral
$$ s(T) := \int\_0^T f(x) \ dx, $$
which is defined to be  

$$ \sigma(T) := \int\_0^T f(x) (1-\frac{x}{T}) \ dx. $$


1. Does the following inequality hold?
$$\operatorname{lim sup}\sigma(T) \leq \operatorname{lim sup} s(T), \quad T \uparrow \infty.$$
2. Suppose $\sigma(T)$ diverges to $\infty$ as $T \uparrow \infty$. Does it follow that $s(T)$ diverges to $\infty$ too?


One expects that the answer to (1) is Yes and the answer to (2) is No but I do not have a proof of the one and a counterexample for the other.



Integral analog of an inequality for the Cesàro mean of a sequence
$\newcommand\si\sigma$Note that
$$\si(T)=\frac1T\,\int\_0^T dx f(x)\,\int\_x^T dt
=\frac1T\,\int\_0^T dt\,\int\_0^t dx\,f(x)
=\frac1T\,\int\_0^T dt\,s(t).$$
Take any real $L>\limsup\_{T\to\infty} s(T)$ (if such $L$ exists, that is, if $\limsup\_{T\to\infty} s(T)<\infty$) and then take any real $A>0$ such that $s(t)\le L$ for all real $t>A$. Then
$$\limsup\_{T\to\infty}\si(T)\le\limsup\_{T\to\infty}\frac1T\,\int\_0^A dt\,s(t)+\limsup\_{T\to\infty}\frac1T\,\int\_A^T dt\,s(t)\le0+L=L,$$
for any real $L>\limsup\_{T\to\infty} s(T)$.


So, the answer to your first question is yes.




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The answer to your second question is no. E.g., suppose that $s(t)=t\sin^2 t$ for real $t\ge0$. Then $\si(T)=\frac{1}{8} \left(2 T^2-2 T \sin (2 T)-\cos (2 T)+1\right)\to\infty$ but $s(T)\not\to\infty$ as $T\to\infty$.


