Let f:[0,1]\to\mathbb{R} be continuous such that
\int_0^1 f(x)\ \mathrm{d}x=1
Determine the minimum possible value of
\int_0^1 (1+x^2)f^2(x)\ \mathrm{d}x
Also, determine the function for which this minimum is attained.
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2 Answers
341This is quite straighforward isnt it? This technique is carried over from algebraic inequalities:
\int_0^1 f^2(x) (1+x^2) \ dx \int_0^1 \frac{1}{1+x^2} \ dx \ge \left( \int_0^1 f(x) \ dx \right)^2
implies that the minimum is \int_0^1 f^2(x) (1+x^2) \ dx \ge \frac{4}{\pi}
From the equality condition for the Schwarz Bunyakovsky and the given condition , we have f(x) = \frac{4}{\pi} \frac{1}{1+x^2} [the other choice is the negative of this function, which is obvously not admissible]
