Prove Integral Inequality: f Nonnegative, Continuous on [0,1]

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SUMMARY

The integral inequality states that for a nonnegative, continuous function \( f \) defined on the interval \([0,1]\), the following holds: \(\left( \int_0^1 f \, dx \right)^2 < \int_0^1 f^2 \, dx\). The discussion highlights the use of upper sums to approach the proof, specifically comparing \(\inf \sum \Delta x_i (M_i(f))^2\) with \(\inf \left[ \sum \Delta x_i M_i(f) \right]^2\). While continuity of \( f \) is emphasized, it is clarified that the inequality holds for any function as long as the integral exists.

PREREQUISITES
  • Understanding of integral calculus, specifically Riemann integrals
  • Familiarity with concepts of upper sums and lower sums
  • Knowledge of properties of continuous functions on closed intervals
  • Basic understanding of inequalities in the context of integrals
NEXT STEPS
  • Study the proof of the Cauchy-Schwarz inequality in the context of integrals
  • Explore the implications of continuity on the existence of integrals
  • Learn about Riemann sums and their application in proving integral inequalities
  • Investigate the role of nonnegative functions in integral calculus
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Mathematics students, particularly those studying real analysis, and educators looking to deepen their understanding of integral inequalities and their proofs.

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Homework Statement


For f nonnegative and continuous on [0,1], prove.
\left( \int f \right) ^2 &lt; \int f^2
With the limits from 0 to 1.



Homework Equations





The Attempt at a Solution


I was trying to use Upper sums, i.e.
\inf \sum \Delta x_i M_i(f^2) = \inf \sum \Delta x_i (M_i(f))^2
and then compare this to \inf \left[ \sum \Delta x_i M_i(f) \right] ^2
Am I in the correct way to prove it?
Why does f is required to be continuous, I didn't use this fact!
 
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Yes, this is the correct way of showing this.

About the continuity of f. You don't really need that here. The inequality is good for any function. But you got to make sure that the integral exists. And that is probably why they chose f to be continuous, because otherwise the integral may not exist...
 

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