Consider the set of positive continuous periodic functions \(f\) on \([0,2\pi]\) satisfying \[ \frac1{2\pi}\int_0^{2\pi} f(\theta)\,d\theta=1. \] What is the largest possible value of \[ \exp\left(\frac1{2\pi}\int_0^{2\pi}\log f(\theta)\,d\theta\right) \] for such functions? Prove that your answer is correct.
Solution
The largest possible value is \[ \boxed{1}. \]
1. Jensen’s inequality gives the upper bound.
Let \[ d\nu(\theta)=\frac{d\theta}{2\pi}, \] so that \(\nu\) is a probability measure on \([0,2\pi]\). Since \(f\) is positive and continuous on the compact interval, \(\log f\) is integrable. The logarithm is concave, so Jensen’s inequality gives \[ \int \log f\,d\nu \le \log\left(\int f\,d\nu\right). \] By the normalization hypothesis, \[ \int f\,d\nu=1, \] hence \[ \frac1{2\pi}\int_0^{2\pi}\log f(\theta)\,d\theta\le0. \] Exponentiating yields \[ \exp\left(\frac1{2\pi}\int_0^{2\pi}\log f(\theta)\,d\theta\right)\le1. \]
2. The bound is attained.
For the constant function \(f\equiv1\), the mean-one constraint holds and \[ \exp\left(\frac1{2\pi}\int_0^{2\pi}\log1\,d\theta\right)=1. \] Thus the maximum is exactly \(1\).
Moreover, because \(\log\) is strictly concave, equality in Jensen’s inequality occurs only when \(f\) is constant almost everywhere. Since \(f\) is continuous, equality occurs exactly for \(f\equiv1\).