Jensen's Inequality using Taylor Expansion¶

Given: \(\sum\limits_{k = 1}^n w_k = 1,f''(x)\ge 0\)

Prove: \(\sum\limits_{k = 1}^n w_kf(x_k)\ge f\left(\sum\limits_{k = 1}^n w_kx_k\right)\)

Proof:

\(\because f''(x)\ge 0\)

\(\Rightarrow f(x)\ge f(x_0) + f'(x_0)(x-x_0)\)

let \(x_0 = \sum\limits_{k = 1}^n w_kx_k\)

\(x\to x_1,x_2,\cdots,x_n\) we have

\(f(x_k)\ge f(x_0) + f'(x_0)(x_k-x_0)\)

\(\Rightarrow w_kf(x_k)\ge w_kf(x_0) + w_kf'(x_0)(x_k-x_0)\)

\(\Rightarrow \sum\limits_{k = 1}^n w_kf(x_k)\ge \sum\limits_{k = 1}^n w_kf(x_0) + \sum\limits_{k = 1}^n w_kf'(x_0)(x_k-x_0)\)

\(\because\sum\limits_{k = 1}^n w_kf'(x_0)(x_k-x_0) = f'(x_0)\left(\sum\limits_{k = 1}^n w_kx_k - x_0\right) = 0\)

\(\therefore \sum\limits_{k = 1}^n w_kf(x_k)\ge f(x_0) = f\left(\sum\limits_{k = 1}^n w_kx_k\right)\)