> [!theorem] > > Let $(\Omega, \cf, \bp)$ be a [[Measure Space|measure space]], $f: \real \to \real$ be a [[Convex Function|convex function]], and $g \in L^1$ be [[Integrable Function|integrable]], then > $ > \int f \circ g \ge f \braks{\int g} > $ > > ### Second Derivatives > > Let $x \in \real$, then > $ > \frac{f(x + h) - f(x)}{h} \quad \frac{f(x - h)- f(x)}{h} > $ > are non-decreasing for $h > 0$. Moreover, > $ > \frac{f(x + h) - f(x)}{h} \ge \frac{f(x) - f(x - h)}{h} > $ > *Proof*. Let $p \in [0, 1]$, then > $ > f(x + ph) \le (1 - p)f(x) + pf(x + h) > $ > by convexity. In which case, > $ > \begin{align*} > f(x + ph) &\le (1 - p)f(x) + pf(x + h) \\ > f(x + ph) - f(x) &\le p(f(x + h) - f(x)) > \end{align*} > $ > so > $ > \frac{f(x + ph) - f(x)}{ph} \le \frac{f(x + h) - f(x)}{h} > $ > On the other hand, > $ > \begin{align*} > f(x - ph) &\le (1 - p)f(x) + pf(x - h) \\ > f(x - ph) - f(x) &\le p(f(x - h) - f(x)) \\ > \frac{f(x - ph) - f(x)}{ph} &\le \frac{f(x - h) - f(x)}{h} > \end{align*} > $ > and > $ > f(x) - f(x - h) \le f(x + h) - f(x) > $ > by convexity. > > ### Tangents > > Define > $ > \begin{align*} > f'_+(x) &= \lim_{h \downto 0}\frac{f(x + h) - f(x)}{h} \\ > f'_-(x) &= \lim_{h \upto 0}\frac{f(x) - f(x - h)}{h} > \end{align*} > $ > and let $c = \int g$, $L(x) = f(c) + a(x - c)$. where $a \in [f'_-(x), f_+'(x)]$. By convexity, $L \le f$. So > $ > \begin{align*} > \int f \circ g &\le \int L \circ g \\ > &= \int \braks{f(c) + a(g - c)} \\ > &=f(c) + a\int\braks{g - \int g} \\ > &= f\braks{\int g} > \end{align*} > $