Lebesgue Integral Computation Involving Infimum

Question

Problem. Suppose $f$ is nonnegative, $\int_{\mathbb{R}^n} f = 1$, and $f(x) \leq 1$ for all $x \in \mathbb{R}^n$. Calculate the following Lebesgue integral

$$\inf \int_{\mathbb{R}^n}\|x\|_2^2f(x)dx,$$

with the infimum taken over all functions $f$.

My idea was to define $f_k(x) := x_k^2f(x)$ such that

$$\int_{\mathbb{R}^n}\|x\|_2^2f(x)dx = \int_{\mathbb{R}^n}(x_1^2 + \dots + x_n^2)f(x) dx= \int_{\mathbb{R}^n} \sum_{k=1}^n f_k.$$

But I don't see a way to use MCT or DCT since the series is finite, so I'm not sure if this is the right approach?

Answer 1

This is what is called The Bathtub Principle (Analysis, by Lieb and Loss). In case the link is not visible the theorem briefly says:

$\textbf{Theorem:}$ If, $(\Omega,\mu)$ is a sigma-finite measure space and $G:\Omega \to \mathbb{R}$ is a measurable functions with sublevels of finite measure (i.e., $\mu(\{G < t\}) < \infty$ for all $t \in \mathbb{R}$) then the infimum of the functional $$J(f) := \int_{\Omega} Gf\,d\mu$$ in the class of functions $\displaystyle \mathscr{C} = \left\{f: 0 \le f \le 1, \int_{\Omega} f\,d\mu = m\right\}$ is attained by $f = \chi_{\{G < s\}} + c\chi_{\{G = s\}}$ where, $s = \sup \{t: \mu(\{G < t\}) \le m\}$ and $c\mu(\{G = s\}) = m - \mu(\{G < s\})$.

(The proof is left as an exercise and I can help you along the way if you need it. The strict monotonicity of $G(x) = |x|^2$ should make the proof much simpler in this case.)