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Fall 2010 - Problem 2

Jensen's inequality

Prove the following form of Jensen's inequality: If $\func{f}{\br{0,1}}{\R}$ is continuous then

\int_0^1 e^{f\p{x}} \,\diff{x} \geq \exp\p{\int_0^1 f\p{x} \,\diff{x}}.

Moreover, if equality occurs then $f$ is a constant function.

Solution.

Since $f$ is continuous, it is in particular integrable, so let $c = \int_0^1 f\p{x} \,\diff{x}$ . Consider $g\p{x} = e^x - \p{e^c + e^c\p{x - c}}$ , i.e., the error between $e^x$ and its tangent line at $x = c$ . Observe that

g'\p{x} = e^x - e^c = 0 \iff x = c \quad\text{and}\quad g''\p{x} = e^x > 0.

Thus, $g$ is convex, so it has a unique global minimum at $x = c$ , which is $g\p{c} = 0$ . This proves that $e^x \geq e^c + e^c\p{x - c}$ for $x \in \R$ . In particular, $e^{f\p{x}} \geq e^c + e^c\p{f\p{x} - c}$ for $x \in \br{0, 1}$ , so integrating both sides yields

\int_0^1 e^{f\p{x}} \,\diff{x} \geq e^c + e^c\p{\int_0^1 f\p{x} \,\diff{x} - c} = \exp\p{\int_0^1 f\p{x} \,\diff{x}},

since the Lebesgue measure on $\br{0, 1}$ has total mass $1$ and because $c = \int_0^1 f\p{x} \,\diff{x}$ by definition.

Observe that in our proof, $e^{f\p{x}} > e^c + e^c\p{f\p{x} - c} \iff g\p{x} > 0$ . By compactness of $\br{0, 1}$ and continuity of $g$ , $g$ attains its minimum $m > 0$ , so the strict inequality holds in integration:

\int_0^1 g\p{x} \,\diff{x} \geq m > 0 \implies \int_0^1 e^{f\p{x}} \,\diff{x} > \exp\p{\int_0^1 f\p{x} \,\diff{x}}.

Thus, equality occurs if and only if $e^{f\p{x}} = e^c + e^c\p{f\p{x} - c} \iff g\p{f\p{x}} = 0$ everywhere. But because $g$ has a unique global minimum at $x = c$ , it follows that $f\p{x} = c$ everywhere, which completes the proof.