Mathematics Department

Math 340 Home, Textbook Contents, Online Homework Home

Warning: MathJax requires JavaScript to process the mathematics on this page.
If your browser supports JavaScript, be sure it is enabled.

THEOREM:

If a continuous function $G(x)$ has the property that $$\int_{a}^{c} G(x)h(x)dx =0$$ for all smooth functions $h(x)$ with $h(a)=0=h(c)$, then $G(x)\equiv 0$ for all $x$ in the interval $(a,c)$.

SKETCH OF PROOF:

The idea here is that if there is a point $x_0$ where $G(x_0)\ne 0$, then there we can find a smooth function $h(x)$ such that $\int_{a}^{c} G(x)h(x)dx \ne 0$. Since we are given the integral is 0 for all choices of $h(x)$, it follows that we must have $G(x) = 0$ at all points.

So to start we assume we have a point $x_0$ with $G(x_0)\ne 0$. We suppose that $G(x_0)>0$ (the proof for the case where $G(x)<0$ is almost identical, just change the words positive and negative where needed). Now since $G(x)$ is continuous, there must be some little interval of $x_0$ where $G(x)>0$ for all points in the interval, say $(x_0-\delta, x_0+\delta)$ as pictured. Note that $G(x)$ might be positive for a ways outside the interval, or it might be negative or 0. We only need to know that $G(x)>0$ for every point in the interval.

Now we pick a "bump function" $h(x)$ which is identically 0 outside the interval $(x_0-\delta,x_0+\delta)$, but is positive for all the points in the interval, as pictured. Then we compute the integral of $G(x)h(x)$. Since $h(x)=0$ for all points outside the interval, $G(x)h(x)=0$ for all points outside the interval, so $$\int_{a}^{c} G(x)h(x)dx = \int_{x_0-\delta}^{x_0+\delta} G(x)h(x)dx $$ (this is why we didn't care what $G(x)$ did outside the interval). On the other hand, we know both $G(x)>0$ and $h(x)>0$ for all points in the interval, so $G(x)h(x)>0$ for all points in the interval $(x_0-\delta,x_0+\delta)$. And if the function is everywhere positive, there is no chance of cancellation so the integral must also be positive, i.e. $$\int_{x_0-\delta}^{x_0+\delta} G(x)h(x)dx > 0.$$ But this contradicts the given condition that the integral is always 0. So the only way to have the integral be 0 for all choices of $h(x)$ is to have $G(x)=0$ for all values of $x$ between $a$ and $c$.

Return to Calculus of Variations


If you have any problems with this page, please contact bennett@math.ksu.edu.
©2010, 2014 Andrew G. Bennett