Proof of the Integral Form of Jensen's Inequality
Theorem
Given a convex function $ \phi : [a,b] \to \mathbb{R}$ and $f: [0,1] \to [a,b]$, if $\phi \circ f$ is integrable over $[0,1]$, then $$ \phi \left( \int_{0}^{1} f(x) dx \right) \le \int_{0}^{1} (\phi \circ f ) (x) dx $$
Explanation
Of course, given the conditions, the integration interval can also be changed through substitution, etc. Unlike finite form, which generalizes the number of terms using definitions, integration form becomes an inequality that the function traverses over the integration symbol.
Proof
By the Mean Value Theorem for Integrals, for some constant $c$, we can set $\displaystyle \int_{0}^{1} f(x) dx = c \in (a,b)$. According to the definition of $c$, $$ \phi \left( \int_{0}^{1} f(x) dx \right) = \phi (c) + s \left( \int_{0}^{1} f(x) dx - c \right) $$ Since the above equation holds for all $s \in \mathbb{R}$, $$\displaystyle s = \sup_{x \in [a,c) } {{\phi (c) - \phi (x)} \over {c -x}}$$ is fine to be defined. For $[c, y] \subset [a,b]$ satisfying $y$, since $\displaystyle {{f(x) - f(c)} \over {x - c}} \le {{f(y) - f(x) } \over {y - x}}$, $$s \le {{\phi (y) - \phi (c)} \over {y- c}} $$ Reorganizing the found equation, $$ \phi (c) + s (y - c) \le \phi (y) $$ Meanwhile, in the case of $[y, c] \subset [a,b]$, $$ s \ge {{\phi (c) - \phi (y)} \over {c - y}} $$ That is, for all $y \in [a,b]$, $\phi (c) + s (y - c) \le \phi (y)$ holds and $[a,b]$ is the range of $f$, so it can be put as $y = f(x)$ to get the following. $$ \phi (c) + s ( f(x) - c) \le \phi ( f(x) ) $$ Taking $\displaystyle \int_{0}^{1}$ to both sides, $$ \int_{0}^{1} \left\{ \phi (c) + s ( f(x) - c) \right\} dx \le \int_{0}^{1} (\phi (f (x) ) dx $$ and organizing yields $$ \phi \left( \int_{0}^{1} f(x) dx \right) + s \int_{0}^{1} f(x) dx - s c \le \int_{0}^{1} (\phi \circ f) (x) dx $$ Lastly, since $\displaystyle \int_{0}^{1} f(x) dx = c$ was true, it is $\displaystyle s \int_{0}^{1} f(x) dx - s c = 0$, and we obtain the following inequality we wanted. $$ \phi \left( \int_{0}^{1} f(x) dx \right) \le \int_{0}^{1} (\phi \circ f) (x) dx $$
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