Calculus through Inequality

This problem involves something that looks like the $L^2$ inner product and it has the word "inequality" in the title, so I figured maybe the solution would involve the Cauchy-Schwarz inequality: $| (\int_0^1 u(x)v(x)dx |^2 \leq \left( \int_0^1 |u(x)|^2dx \right) \left( \int_0^1 |v(x)|^2dx \right)$ . We're looking for a minimum, so let's try to arrange for the integral $I(f) = \int_0^1 (1+x^2)(f(x))^2dx$ to be one of the terms on the larger right-hand side of the inequality, say $u(x)^2 = (1+x^2)(f(x))^2$ . Then $u(x) = \sqrt{1 + x^2}f(x)$ . If we choose $v(x) = \frac{1}{\sqrt{1 + x^2}}$ , then the left-hand side of the inequality would just be $\int_0^1 f(x) dx$ . So: $\begin{aligned} 1 &= \int_0^1 f(x)dx \\ &= \int_0^1 \left(\sqrt{1+x^2}f(x) \right) \left( \frac{1}{\sqrt{1+x^2}}\right)dx \\ &\leq \left( \int_0^1 (1+x^2)(f(x))^2 dx \right) \left( \int_0^1 \frac{dx}{1+x^2} \right) \\ &= I(f) (\tan^{-1}(x)|_0^1) \\ &= I(f) \frac{\pi}{4} \end{aligned}$

Therefore, $I(f) \geq \frac{4}{\pi}$ . We have found a lower bound for values of $I(f)$ over $A$ , but is it the minimum, i.e, the greatest lower bound? Fortunately, the work we've already done suggests a candidate for a function $f \in A$ for which $I(f)$ attains this lower bound. Namely, $f(x) = \frac{4}{\pi} \frac{1}{1+x^2}$ . It's easy to check that $f \in A$ and $I(f) = \frac{4}{\pi}$ .