version of the fundamental lemma of calculus of variations

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

$${\int }_a^{b}f(x)\phi (x)𝑑x=\mathrm{\hspace{0.33em}0}$$

for all functions $\phi$ continuously differentiable on the interval and vanishing at its end points, then $f(x)\equiv \mathrm{\hspace{0.17em}0}$  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  $f(x_0)\ne 0$;  for example  $f(x_0)>0$.  The continuity of $f$ implies that there are the numbers $\alpha$ and $\beta$ such that    and  $f(x)>0$  for all $x\in [\alpha ,\beta ]$.  Now the function ${\phi }_0$ defined by

${\phi }_0(x):=\{\begin{array}{cc}{(x-\alpha )}^2{(x-\beta )}^2\hspace{1em}\text{for}\alpha \leqq x\leqq \beta ,\hfill & \\ 0\hspace{1em}\hspace{1em}\text{otherwise}\hfill & \end{array}$

fulfils the requirements for the functions $\phi$.  Since both $f$ and ${\phi }_0$ are positive on the open interval  $(\alpha ,\beta )$,  we however have

$${\int }_a^{b}f(x){\phi }_0(x)𝑑x={\int }_{\alpha }^{\beta }f(x){\phi }_0(x)𝑑x>\mathrm{\hspace{0.33em}0}.$$

Thus the antithesis causes a contradiction.  Consequently, we must have  $f(x)\equiv \mathrm{\hspace{0.17em}0}$.