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$\newcommand\si\sigma$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 $L>\limsup_{T\to\infty} s(T)$ 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 $L>\limsup_{T\to\infty} s(T)$.

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 $T\to\infty$.

$\newcommand\si\sigma$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 $T\to\infty$.

$\newcommand\si\sigma$Note that $$\si(T)=\frac1T\,\int_0^T dt\,s(t).$$ Take any $L>\limsup s(T)$ and then take any real $A>0$ such that $s(t)\le L$ for all real $t>A$. Then $$\limsup\si(T)\le\limsup\frac1T\,\int_0^A dt\,s(t)+\limsup\frac1T\,\int_A^T dt\,s(t)\le0+L=L,$$ for any $L>\limsup s(T)$.

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 $T\to\infty$.

$\newcommand\si\sigma$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 $L>\limsup_{T\to\infty} s(T)$ 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 $L>\limsup_{T\to\infty} s(T)$.

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 $T\to\infty$.

$\newcommand\si\sigma$Note that $$\si(T)=\frac1T\,\int_0^T dt\,s(t).$$ Take any $L>\limsup s(T)$ and then take any real $A>0$ such that $s(t)\le L$ for all real $t>A$. Then $$\limsup\si(T)\le\limsup\frac1T\,\int_0^A dt\,s(t)+\limsup\frac1T\,\int_A^T dt\,s(t)\le0+L=L,$$ for any $L>\limsup s(T)$.

So, the answer to your first question is yes.

$\newcommand\si\sigma$Note that $$\si(T)=\frac1T\,\int_0^T dt\,s(t).$$ Take any $L>\limsup s(T)$ and then take any real $A>0$ such that $s(t)\le L$ for all real $t>A$. Then $$\limsup\si(T)\le\limsup\frac1T\,\int_0^A dt\,s(t)+\limsup\frac1T\,\int_A^T dt\,s(t)\le0+L=L,$$ for any $L>\limsup s(T)$.

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 $T\to\infty$.