version of the fundamental lemma of calculus of variations

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

 $\int_{a}^{b}\!f(x)\varphi(x)\,dx\;=\;0$

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

Proof.  We make the antithesis that $f$ does not vanish identically.  Then there exists a point $x_{0}$ of the open interval  $(a,\,b)$  such that  $f(x_{0})\neq 0$;  for example  $f(x_{0})>0$.  The continuity of $f$ implies that there are the numbers $\alpha$ and $\beta$ such that  $a<\alpha  and  $f(x)>0$  for all  $x\in[\alpha,\,\beta]$.  Now the function $\varphi_{0}$ defined by

 $\displaystyle\varphi_{0}(x)\;:=\;\begin{cases}(x\!-\!\alpha)^{2}(x\!-\!\beta)^% {2}\quad\mbox{for}\;\;\alpha\leqq x\leqq\beta,\\ 0\qquad\mbox{otherwise}\end{cases}$

fulfils the requirements for the functions $\varphi$.  Since both $f$ and $\varphi_{0}$ are positive on the open interval  $(\alpha,\,\beta)$,  we however have

 $\int_{a}^{b}\!f(x)\varphi_{0}(x)\,dx\;=\;\int_{\alpha}^{\beta}\!f(x)\varphi_{0% }(x)\,dx\;>\;0.$

Thus the antithesis causes a contradiction.  Consequently, we must have  $f(x)\,\equiv\,0$.

Title version of the fundamental lemma of calculus of variations VersionOfTheFundamentalLemmaOfCalculusOfVariations 2013-03-22 19:12:04 2013-03-22 19:12:04 pahio (2872) pahio (2872) 4 pahio (2872) Theorem msc 26A15 msc 26A42 fundamental lemma of calculus of variations EulerLagrangeDifferentialEquation