version of the fundamental lemma of calculus of variations

Lemma.  If a real function f is continuousMathworldPlanetmathPlanetmath on the interval[a,b]  and if

abf(x)φ(x)𝑑x= 0

for all functions φ continuously differentiable on the interval and vanishing at its end points, then  f(x) 0  on the whole interval.

Proof.  We make the antithesis that f does not vanish identically.  Then there exists a point x0 of the open interval  (a,b)  such that  f(x0)0;  for example  f(x0)>0.  The continuity of f implies that there are the numbers α and β such that  a<α<x0<β<b  and  f(x)>0  for all  x[α,β].  Now the function φ0 defined by

φ0(x):={(x-α)2(x-β)2forαxβ,0  otherwise

fulfils the requirements for the functions φ.  Since both f and φ0 are positive on the open interval  (α,β),  we however have

abf(x)φ0(x)𝑑x=αβf(x)φ0(x)𝑑x> 0.

Thus the antithesis causes a contradictionMathworldPlanetmathPlanetmath.  Consequently, we must have  f(x) 0.

Title version of the fundamental lemma of calculus of variationsMathworldPlanetmathPlanetmath
Canonical name VersionOfTheFundamentalLemmaOfCalculusOfVariations
Date of creation 2013-03-22 19:12:04
Last modified on 2013-03-22 19:12:04
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 4
Author pahio (2872)
Entry type Theorem
Classification msc 26A15
Classification msc 26A42
Synonym fundamental lemma of calculus of variations
Related topic EulerLagrangeDifferentialEquation