Proof of the Integral Form of Jensen's Inequality
Theorem
For a convex function $ \phi : [a,b] \to \mathbb{R}$ and $f: [0,1] \to [a,b]$, if $\phi \circ f$ is integrable on $[0,1]$, then $$ \phi \left( \int_{0}^{1} f(x) dx \right) \le \int_{0}^{1} (\phi \circ f ) (x) dx $$
Explanation
Obviously, as long as the given conditions are satisfied, the interval of integration can also be changed through substitution and the like. Unlike the finite form, which generalizes the number of terms using the definition, the integral form becomes an inequality in which a function moves across the integral sign.
Proof
By the mean value theorem for integrals, for some constant $c$ we may set $\displaystyle \int_{0}^{1} f(x) dx = c \in (a,b)$. By the definition of $c$, $$ \phi \left( \int_{0}^{1} f(x) dx \right) = \phi (c) + s \left( \int_{0}^{1} f(x) dx - c \right) $$ The above equality holds for all $s \in \mathbb{R}$, so we may define $$\displaystyle s = \sup_{x \in [a,c) } {{\phi (c) - \phi (x)} \over {c -x}}$$ without any problem. For $y$ satisfying $[c, y] \subset [a,b]$, since $\displaystyle {{f(x) - f(c)} \over {x - c}} \le {{f(y) - f(x) } \over {y - x}}$, we have $$s \le {{\phi (y) - \phi (c)} \over {y- c}} $$ Rearranging the equation obtained above, we get $$ \phi (c) + s (y - c) \le \phi (y) $$ On the other hand, in the case $[y, c] \subset [a,b]$, $$ s \ge {{\phi (c) - \phi (y)} \over {c - y}} $$ That is, $\phi (c) + s (y - c) \le \phi (y)$ holds for all $y \in [a,b]$, and since $[a,b]$ is the range of $f$, we may set $y = f(x)$ and obtain the following. $$ \phi (c) + s ( f(x) - c) \le \phi ( f(x) ) $$ Applying $\displaystyle \int_{0}^{1}$ to both sides, $$ \int_{0}^{1} \left\{ \phi (c) + s ( f(x) - c) \right\} dx \le \int_{0}^{1} (\phi (f (x) ) dx $$ and rearranging yields $$ \phi \left( \int_{0}^{1} f(x) dx \right) + s \int_{0}^{1} f(x) dx - s c \le \int_{0}^{1} (\phi \circ f) (x) dx $$ Finally, since $\displaystyle \int_{0}^{1} f(x) dx = c$, we have $\displaystyle s \int_{0}^{1} f(x) dx - s c = 0$, and we obtain the following inequality that we wanted. $$ \phi \left( \int_{0}^{1} f(x) dx \right) \le \int_{0}^{1} (\phi \circ f) (x) dx $$
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