Consider a function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that it is continuous.
For any $t \in \mathbb{R}$, let $$I(t) := \frac{\int_{-t}^{t}(f(s))^2ds}{t}$$ and let $$I_0 := \lim_{t \to \infty} \frac{\int_{-t}^{t}(f(s))^2ds}{t}$$
From mean value theorem for integrals, $I(t) = 2(f(c))^2$ for some $c \in (-t,t)$. Since $I_0 = \lim_{t \to \infty}I(t)$ can I say $I_0 = 2(f(c_0))^2$ for some $c_0 \in \mathbb{R}$? I am confused in the step of application of the limit, since mean value theorem requires a compact interval, so can we simply extend this statement to the whole of $\mathbb{R}$?
Take $f(s) = s.$ Then $$ I(t) = \frac{1}{t}\int_{-t}^t f(s)^2 \,ds = \frac{1}{t}\int_{-t}^t s^2 \, ds = \frac{1}{t} \left[\frac{s^3}{3}\right]_{s=-t}^t= \frac{2}{3}t^2. $$ So for any fixed $t,$ we can put $$c = \frac{t}{\sqrt{3}} \in (-t,t)$$ which satisfies $$ 2f(c)^2 = 2c^2 = \frac{2}{3}t^2 = I(t). $$ But $$ I_0 = \lim_{t\to \infty} I(t) = \lim_{t\to \infty} \frac{2}{3}t^2 = +\infty, $$ so there exists no $c_0\in \mathbb{R}$ such that $$ 2f(c_0)^2 = I_0. $$