Convergence of integrals

For I, suppose that the relevant limits exist. Then the relevant limits are equal to $0$ as otherwise it cannot be the case that $\int_{-\infty}^\infty |f(x)|\,\mathrm{d}x < \infty.$ As a result, there exists $R > 0$ such that for $|x| > R$ , $|f(x)| < 1$ . Now note that for $|x| > R$ , $(f(x))^2 = |f(x)|^2 < |f(x)|$ , so $\int_{\mathbb{R}\backslash[-R,R]} (f(x))^2 \,\mathrm{d}x \leq \int_{\mathbb{R}\backslash[-R,R]} |f(x)| \,\mathrm{d}x < \infty.$ Furthermore, as $f$ is continuous, $f$ is bounded on the interval $[-R,R]$ . As a result, $\int_{-R}^R (f(x))^2\,\mathrm{d}x \leq 2R \cdot \left(\|f\|_{[-R,R]}\right)^2 < \infty.$ Hence $\int_{-\infty}^\infty (f(x))^2\,\mathrm{d}x = \int_{-R}^R (f(x))^2\,\mathrm{d}x + \int_{\mathbb{R}\backslash[-R,R]} (f(x))^2\,\mathrm{d}x < \infty.$

For II, a single counterexample is sufficient. For example, note that $g_1, g_2: \mathbb{R} \to \mathbb{R}, g_1(x) = \sin(x^2), g_2(x) = \cos(x^2)$ are infinitely differentiable, while $\int_{-\infty}^\infty \sin(x^2) \,\mathrm{d}x =\int_{-\infty}^\infty \cos(x^2) \,\mathrm{d}x = \sqrt{\frac{\pi}{2}} < \infty,$ but $\int_{-\infty}^\infty (\sin^2(x^2) + \cos^2(x^2)) \,\mathrm{d}x = \int_{-\infty}^\infty 1 \,\mathrm{d}x = \infty,$ implying that at least one of the integrals $\int_{-\infty}^\infty \sin^2(x^2) \,\mathrm{d}x \text{ and } \int_{-\infty}^\infty \cos^2(x^2) \,\mathrm{d}x$ cannot be finite as both integrands are non-negative for any $x \in \mathbb{R}$ .