How does Jensen's inequality apply to this problem?

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SUMMARY

This discussion focuses on applying Jensen's inequality to demonstrate that for an integrable function \( f \) over the interval \([0,1]\), the inequality \(\exp\left[\int_0^1 f(x)\,dx\right] \leq \int_0^1\exp(f(x))\,dx\) holds true. The exponential function is identified as convex, which is critical in deriving the inequality. The solutions provided by Ackbach and Opalg illustrate the steps taken to integrate the inequality and confirm its validity through the properties of convex functions.

PREREQUISITES
  • Understanding of Jensen's inequality in the context of convex functions.
  • Knowledge of integrable functions and their properties over closed intervals.
  • Familiarity with the exponential function and its convexity.
  • Basic integration techniques and properties of definite integrals.
NEXT STEPS
  • Study the properties of convex functions and their implications in inequality proofs.
  • Explore advanced applications of Jensen's inequality in various mathematical contexts.
  • Learn about the implications of convexity in optimization problems.
  • Investigate other inequalities related to integrals, such as Hölder's and Minkowski's inequalities.
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Mathematicians, students studying real analysis, and anyone interested in the applications of inequalities in mathematical proofs will benefit from this discussion.

Chris L T521
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Here's this week's problem!

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

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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!
 
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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]
 

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