How does Jensen's inequality apply to this problem?

  • Thread starter Thread starter Chris L T521
  • Start date Start date
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
Chris L T521
Gold Member
MHB
Messages
913
Reaction score
0
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!
 
Physics news on Phys.org
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]