Integral analog of an inequality for the Cesàro mean of a sequence

Question

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.

Answer 1

$\newcommand\si\sigma$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.

---

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 $T\to\infty$.