The following is my proof for the title:
We have to show that a strongly convex function $f$ meets the following $$f(\lambda x + (1-\lambda)y)<\lambda f(x) + (1-\lambda)f(y). $$
By the definition of the strongly convex function $f$:
\begin{align*}
f(y)\geq f(x) + \langle \nabla f(x),y-x\rangle + \frac{m}{2}\|y-x\|_2^2,
\end{align*}where $m>0$. Consider both sides of the definition of strictly convex $f$:
- The right-hand side: \begin{align*} &\lambda f(x)+(1-\lambda)f(y) \\ &\geq \lambda f(x) + (1-\lambda) \bigg[f(x) + \langle \nabla f(x),y-x\rangle + \frac{m}{2}\|y-x\|_2^2 \bigg] \\ & = f(x) + (1-\lambda)\langle \nabla f(x),y-x\rangle + \frac{m}{2}(1-\lambda) \|y-x\|_2^2 \end{align*}
- The left-hand side: \begin{align*} &f(\lambda x + (1-\lambda)y) \\ &\geq f(x) + \langle \nabla f(x),(\lambda x + (1-\lambda)y)-x\rangle + \frac{m}{2}\|(\lambda x + (1-\lambda)y)-x\|_2^2 \\ &= f(x) + (1-\lambda)\langle \nabla f(x),y-x\rangle + \frac{m}{2}(1-\lambda)^2\|y-x\|_2^2 \end{align*}
Since $\lambda \in (0,1)$, $(1-\lambda)>(1-\lambda)^2$; therefore, we prove that $\lambda f(x)+(1-\lambda)f(y) > f(\lambda x + (1-\lambda)y)$
My question: There is no evidence show that this is a tight lower bound.
EX: $10>9$ and $20>1$, we cannot say that $(20-10) > (1 - 9)$. How to fix this problem?