Prove Jensen's inequality for the exponential function on [0,1]

[Chris L T521]

Here's this week's problem!

-----

Problem: Let $f$ be integrable over $[0,1]$. Show that
\[\exp\left[\int_0^1 f(x)\,dx\right] \leq \int_0^1\exp(f(x))\,dx.\]

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

This week's problem was correctly answered by Ackbach and Opalg. You can find Opalg's solution below.

[sp]This is a particular case of Jensen's inequality.

The exponential function is convex, so the tangent at any point lies below the curve (except at the point where they touch). The equation of the tangent at the point $(t,e^t)$ is $y = e^tx + (1-t)e^t.$ It follows that $e^x \geqslant e^tx + (1-t)e^t. \quad(*)$

Let $$J = \int_0^1 \!\!f(x)\,dx.$$ Take $t=J$ in (*) (and replace $x$ by $f(x)$) to see that $ \exp(f(x)) \geqslant e^Jf(x) + (1-J)e^J.$

Now integrate that from $0$ to $1$: $$\int_0^1\!\! \exp(f(x))\,dx \geqslant e^J\!\!\int_0^1\!\!f(x)\,dx + \int_0^1\!\!(1-J)e^J\,dx = e^JJ + (1-J)e^J = e^J = \exp\left[ \int_0^1 \!\!f(x)\,dx\right].$$[/sp]