The question below is inspired by this one.

Assume $f$ is continuous on $[a,b]$ and $\int_a ^b f(x)g(x)dx=0 $ for every function $g$ possessing continuous derivative such that $g(a)=g(b)=0$. Is it true that $f\equiv 0$ on the interval $[a,b]$?

The hint I was given is to cook up a smooth function with compact support out of $e^{-1/x^2}$. Expanded version of the hint was this: set $f(x)=e^{-1/x^2}$ for $x$ positive and $f(x)=0$ for $x$ non-positive and consider $\int_0^xf(t)f(1-t)dt$.

I have no idea what lies behind this hint (i.e., how could one come up with such an integral) and do not know how to use the hint either.


2 Answers 2


The function $h(x) = f(x) f(1-x) $ looks a little like an upside-down bathtub: it's 0 for $x \le 0$ and for $x \ge 1$, but nonzero (indeed, positive) for $0 < x < 1$.

For any interval $[p, q]$, the function $h_{p, q} (x) = f(\frac{x-p}{q-p})$ is similar: it's nonzero on the interior of the interval, zero outside it.

If you scale it up by a constant, you can arrange that $\int_p^q h_{p,q}(x) = 1$.

You'd like to show that for any $c$ in $[a, b]$, $f(c) = 0$.

Suppose not .. suppose that $f(c) = A > 0$. Then there's an interval $[p, q]$, containing $c$, indeed, with $a < p < c < q< b$ on which $f(x) \ge A/2$. (Why?)

That tells you that $$ \int_a^b f(t) h_{p,q}(t)~dt = \int_p^q f(t) h_{p,q}(t) dt \ge \int_p^q \frac{A}{2} h_{p, q}(t) ~ dt. $$ What can you say about that last integral? How does this related to the claim about $f$ and "any $g$"?

  • $\begingroup$ We want to show that $f(c)=0$ for any $c$ in $[a,b]$, not in $[0,1]$, right? If so, then assuming $f(c)=A>0$ for $c\in [a,b]$ we have $f(c)=A>0$ on $[\alpha,\beta]\subset [a,b]$ by continuity and the triangle inequality, so $$\int_a^bf(t)h_{\alpha,\beta}(t)dt\ge \int_{\alpha}^{\beta}\frac{A}{2}h_{\alpha,\beta}(t)dt>0$$ since $\int_{\alpha}^{\beta}h_{\alpha,\beta}(t)dt$ is positive on $[\alpha,\beta]$. Thus for the function $g(x)=h_{\alpha,\beta}(x)$, the integral is positive, which is a contradiction. Is that correct? $\endgroup$
    – user557
    Nov 15, 2017 at 3:42
  • $\begingroup$ Oh, darn. I read the problem as being on $[0, 1]$ rather than $[a, b]$, and then used "a" and "b" for a different thing in my answer. I've edited to make it work for the problem as stated. And your contradiction does indeed look correct, yes. $\endgroup$ Nov 15, 2017 at 4:05
  • $\begingroup$ Thank you. What for do you need to scale $h_{p,q}$ (i.e., why do you need the integral of this function to be one)? Could I replace $e^{-1/x^2}$ with $e^{-1/x}$? $\endgroup$
    – user557
    Nov 15, 2017 at 17:54
  • $\begingroup$ I'd like the last integral to be at least $A/2$; for that to happen, I need the integral of the "$h$ part to be $1$. As for replacing with $e^{-1/x}$...I think that's OK. The one I used has the property that it's actually $C^\infty$ (i.e., has continuous derivatives of all orders), but I see that your test-function $g$ is required only to have continuous first derivative, so mine was overkill. I haven't checked that yours DOES have continuous first deriv, though, so you'd have to confirm that. $\endgroup$ Nov 16, 2017 at 11:39

Suppose that $f(y) \neq 0$ for some $y \in [a,b]$, without loss of generality, $f(y) > 0$. Then for some $\delta > 0$, $f(x) > 0$ for $x \in (y-\delta, y+\delta)$, and assume that $\delta$ is such that this interval is still contained in $[a,b]$ (if not, we are free to make it smaller).

Let $G(x) = \exp\left( \frac{1}{1-x^2} \right)$ for $x \in (-1,1)$ and $G(x)=0$ for $|x| > 1$. Then we make $g(x) = G(\frac{x-y}{\delta})$. Now we have a compactly supported function $g$ whose support is exactly $(y-\delta, y+\delta)$, and $g$ is strictly positive on $(y-\delta, y+\delta)$.

What can we say about $$\int_a^b f(x)g(x) dx = \int_{y-\delta}^{y+\delta} f(x)g(x) dx?$$


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