# Show that $g(a) = g(b) = 0,\ \int_a^b f(x)g(x)dx=0$ implies $f(x)=0$

Suppose $f$ is continuous on $[a, b]$, if for every continuous function $g$ on $[a, b]$ with $g(a) = g(b) = 0, \int_{a}^{b}f(x)g(x) dx = 0$, Show $f(x) = 0, \forall x \in [a, b]$,

I want to prove by contradiction, and then find a continuous $g$ such that $g(a) = g(b) = 0$ but $\int_{a}^{b}f(x)g(x) dx \neq 0$

Proof: Suppose by contradiction that $f(x) > 0$ for some $x_0 \in [a, b]$. Since $f$ is continuous, $\exists \delta$ such that $f(x) > 0, \forall x \in [x_0 - \delta, x_0 + \delta]$. Take $g(x) = \begin{cases} 0 & \text{ if } x \in (a, x_0 - \delta) \cup (x_0+\delta, b) \\ -(x - x_0 - \delta)(x - x_0 + \delta) & \text{ if } x \in (x_0 - \delta, x_0 + \delta) \end{cases}$

Now I want to show that since $g$ is continuous on $[a, b]$, then it is integrable. Thus $\int_{a}^{b} g(x) dx = sup L(g, p)$. However, since the lower sum is $> 0$, it follows that the supremum is also $> 0$. Therefore $\int_{a}^{b} > 0$. A contradiction. However, I do not know how to show that $L(g, p) > 0$

• The negation of $f(x) =0 \forall x$ is not $f(x) \neq 0 \forall x$. – N. S. Apr 6 '15 at 0:35
• It should be for $\exists x \in [a, b]$. Thanks for pointing that out. – Adrian Apr 6 '15 at 0:37
• You cannot take $g = f$ because it might not be the case that $f(a) = f(b) = 0$. My hint: Suppose $f(x) > 0$ (say; similar if it were negative) for some $x \in [a, b]$. Then $f$ is positive in a tiny interval around $x$. Let $g$ be a kind of "bump" function that is positive near x but is 0 outside this interval. What can you conclude? – Pedro Apr 6 '15 at 0:43
• @Pedro, yes that was what my professor suggested as well. I am having trouble coming up with a "bump" function. Could you give me a hint? – Adrian Apr 6 '15 at 0:57
• @Pedro, I came up with a "bump" function. And I edited my original post. However, I am stuck at showing how the lower sum has to be > 0 – Adrian Apr 8 '15 at 2:20

Let $g(x)=f(x)(x-a)(b-x)$ Then $$\int_a^b f(x)^2 (x-a)(b-x) dx =0$$ Note that $$f(x)^2 (x-a)(b-x)\geq 0,\ (x-a)(b-x) > 0\ (x\in (a,b))$$

If for some $x\in [a,b]$, $f(x)\neq 0$ then since $f$ is continuous, there exists a closed set $x\in [s,t]$ : $$f(x)^2\geq c > 0\ {\rm on}\ [s,t] \subseteq [a,b]$$

Hence $$\int_a^b f(x)^2 (x-a)(b-x) dx \geq \int_s^t c (x-a)(b-x) dx > 0$$ So contradiction.

• This beats me. Very nice choice of $g(x)$. +1. – Paramanand Singh Apr 11 '15 at 6:16
• I like solutions that are more clever than my patchwork. – r12 Apr 11 '15 at 6:38
• Could you comment on how you showed that the the above integral is 0? If I understood the question, the integral being 0 is a property that is required. – Kayle of the Creeks Apr 26 '15 at 13:41
• It is an assumption of $f$. If $f$ satisfies such assumption, we must show that $f=0$. That is, when $f$ satisfies the assumption, if $f$ is not $0$, we complete the proof by showing a contradiction. – HK Lee Apr 26 '15 at 13:53

Elaborating on N. S.'s comment, if $f$ is non zero at some $x_0 \in [a,b]$, then it is non-zero in some region $[x_0- \delta , x_0 + \delta ]$. WLOG take $f > 0$ in this region.

Then find a $g$ s.t. $g(a) = g(b) = 0$ but $g(x) > 0$ $\forall x \in [x_0-\delta,x_0+\delta]$ (think of defining $g$ piecewise).

Further, make $g$ non zero only in this region (i.e. $g \geq 0$ everywhere, but $g >0$ only in $[x_0-\delta,x_0+\delta]$.

If $x-\delta < a$ or $x+ \delta >b$, then we take a smaller value for $\delta$. The final part is to show the integral is non-zero, but this should not be hard to do (think finding the area or in terms of step functions).

• One pedantic issue in your construction of g, it can't just be "non-zero" only in that region. For example, if g = -f on [x0 - d, x0 + d], then the integral is zero on [x0 -d, x0 + d], and thus everywhere. – r12 Apr 10 '15 at 23:34
• Very true. I had in the line before stated that $g(x) > 0$, but for clarity I should have made that clear with my use of "Better yet". Having said that, g = -f doesn't work, as it gets a strictly negative integral, but some function that crosses the x-axis could of course work. – Krishan Bhalla Apr 11 '15 at 4:44
• Oh, you're right. g=-f doesn't work. Also, I thought when you wrote "non-zero", you were letting it be implied that it couldn't cross the x-axis from meaning "non-zero everywhere" and intermediate value theorem. – r12 Apr 11 '15 at 6:33

Suppose there exists some $n$ such that $f(n) = a$, $a \neq 0$. Since $f$ is continuous, let $E$ be the open interval around $n$ where $E \subset [a,b]$.

Let $E1$ be the open interval around $n$ such that $sgn(e) = sgn(n) \forall e\in E1 \subset E$.

Let $E2 \subset E1$.

Let $E2 = (c_2, d_2)$, $E1 = (c_1, d_1)$.

Construct $g$ so that $g(x) = f(x)$ for $x \in E1$, $g(x) = 0$ for $[a,b] \cap E2'$, and $g(x)$ connects the parts continuously on $E1 \cap E2'$.

Once you have that, then it should be trivial. Your integrand will be all one sign on $E1$ and non-zero, and zero everywhere else, which means your integral will be non-zero.

EDIT:

I'll go ahead and attempt to construct $g(x)$ on $[c_1,c_2]$ and $[d_2, d_1]$ explicitly.

$g(x) = f(x)* |\frac{x-c_1}{c_2-c_1}|$ on $x \in [c_1,c_2]$, and $g(x) = f(x)*|\frac{x-d_1}{d_2-d_1}|$.

Counter example: $f(x)= const$, $g(x)=sin(x)$, $a=0, b=2 \pi$.

Direct computations shows that $\int_{0}^{2\,\pi }f(x) \; \mathrm{sin}\left( x\right) dx =0$.

So you need more constraints.

I'll rename the $x_0$ you defined to $c \in [a, b]$.

Let $p$ be the partition $(x_0 < x_1 < x_2 < x_3) = (a < c - \frac{\delta}{2} < c + \frac{\delta}{2} < b)$.

On the intervals $[x_0, x_1]$ and $[x_2, x_3]$, either $g$ vanishes or both $f$ and $g$ are positive, so $$\inf_{x \in [x_0, x_1]} f(x)g(x) = \inf_{x \in [x_2, x_3]} f(x)g(x) = 0.$$

On the interval $[x_0 - \frac{\delta}{2}, x_0 + \frac{\delta}{2}]$, both $f$ and $g$ are strictly positive, so $$\inf_{x \in [x_1, x_2]} f(x)g(x) > 0.$$ Summing over all subintervals gives $$L(fg, p) = \delta \inf_{x \in [x_1, x_2]} f(x)g(x) > 0$$

so necessarily $\sup_p L(fg, p)$ > 0.