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As in the title with a bit more detail, assume $f$ is Riemann Integrable from $[0,1] \rightarrow \mathbb R$, prove that $\int_0^1 \exp(f(x))dx \geq \exp(\int_0^1 f(x)\,dx$, I did a bit of research and noticed that this is an example of Jensen's inequality. But how would you prove this without measure theory/purely from the Riemann Integral perspective?

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  • $\begingroup$ Please fix your incorrect title. $\endgroup$ – David G. Stork May 27 at 22:16
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    $\begingroup$ Jensen’s inequality will often be proven in a measure/probability theory course but it doesn’t really rely on any measure theory. The fact that the limits of integration are a distance one away from eachother is important though. $\endgroup$ – Josh Messing May 27 at 22:30
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    $\begingroup$ Jensen's inequality is not hard to prove with Riemann sums. It's a generalization of the definition of convexity. Convexity says that if $a,b \geq 0 $ and $a+b = 1$then $f(ax + by) \leq af(x) + bf(y)$. Generalize this to: if $a_1, \ldots, a_n \geq 0$ and $\sum_i a_i = 1$ then $f(\sum_i a_i x_i) \leq \sum_i a_i f(x_i)$. In slogan form, $f$(weighted average) $\leq$ weighted average($f$) for any convex $f$. Now take $n \rightarrow \infty$ and turn things into Riemann sums. And note that $x \mapsto e^x$ is convex. $\endgroup$ – Jair Taylor May 27 at 22:31
  • $\begingroup$ Have a look at this version. $\endgroup$ – rtybase May 27 at 22:35
  • $\begingroup$ More directly, since $x\mapsto\exp(x)$ is convex, $\exp(f(x))\ge\exp(h)+\exp(h)(f(x)-h)$; specialize to $h=\int_0^1f(x)dx$ and integrate. $\endgroup$ – kimchi lover May 27 at 22:36
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If $\phi$ is convex (as is $\exp$) we have $\phi(t) \ge \phi(t_0) + \phi'(t_0) (t-t_0)$.

Hence $\phi(f(x)) \ge \phi(t_0) + \phi'(t_0) (f(x)-t_0)$ (cf. subdifferential).

Now integrate to get $\int_0^1 \phi(f(x)) dx \ge \phi(t_0) + \phi'(t_0)(\int_0^1 f(x)dx - t_0)$.

Now choose $t_0 = \int_0^1 f(x)dx$ to get $\int_0^1 \phi(f(x)) dx \ge \phi(\int_0^1 f(x)dx)$.

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