# Why does $\int_a^b f(x)h'(x) \, \mathrm{d}x=0$ imply that $f$ is constant?

As an assignment, I have to prove the following:

If $$f(x)$$ is a piecewise continuous function and $$\int_a^b f(x)h'(x) \, \mathrm{d}x=0$$ for all piecewise continuously differentiable $$h(x)$$ that satisfy $$h(a)=h(b)=0$$, then $$f$$ is constant on $$[a,b]$$.

The assignment also provides some hints:

Define $$c:=\frac{1}{b-a} \int_a^b f(x) \, \mathrm{d}x=\frac{1}{b-a} \sum_{i=1}^{m} \left ( \int_{x_{I-1}}^{x_i} f(x) \, \mathrm{d}x \right )$$ and use $$h(x)=\int_a^x f(s)-c \, \mathrm{d}s$$ Show that $$\int_a^b \left ( f(x)-c \right )h'(x) \, \mathrm{d}x=0$$ as well as $$\int_a^b \left ( f(x)-c \right )h'(x) \, \mathrm{d}x = \int_a^b \left ( f(x)-c \right )^2 \, \mathrm{d}x$$ and use this to conclude that $$f(x)-c=0$$ for all $$x \in [a,b]$$.

Now, from what I understand, $$h(b)=0$$: $$h(x)=\int_a^b f(s)-c \, \mathrm{d}s=\int_a^b f(s) \, \mathrm{d}s-\left . cs \right |_{s=a}^b=(b-a)c-(cb-ca)=0$$ Obviously, $$h(a)=0$$ too.
What I don't understand, is how I'm supposed to manipulate $$\int_a^b \left ( f(x)-c \right )h'(x) \, \mathrm{d}x$$ to obtain $$\int_a^b \left ( f(x)-c \right )^2 \, \mathrm{d}x$$. I've tried numerous things (including integration by parts, which looks promising), but to no avail.

• Should the definition of $h(x)$ have an $x$ in it somewhere? – Greg Martin Feb 15 '19 at 18:10
• Right, if the definition were $h(x) = \int_a^x [f(s) - c] ds$, then this all follows through. – Tom Chen Feb 15 '19 at 19:07
• I made sure to have another good look at the assignment. It definitely says $\int_a^b$. I thought it was weird how a function of $x$ did not 'use' $x$, but I wrote it off because this is perfectly possible. I'll ask the professor if this is a typo. – JoieNL Feb 15 '19 at 21:30
• Alright, the assignment is due tomorrow and I still haven't received a reply from the professor, so I'll just assume it's a typo. I'll edit the question and accept the best answer for future reference. – JoieNL Feb 17 '19 at 16:35

Seems to me like a step to prove later the Euler-Lagrange equation in the calculus of variation. Now let us assume $$\int_a^b f(x)h'(x) \, \mathrm{d}x=0$$ for all piecewise continuously differentiable $$h(x)$$ that satisfy $$h(a)=h(b)=0$$

We now choose a special $$h$$ (having the properties above) to show that $$f$$ must be a constant and set $$h(x)=\int_a^x f(s)-c \, \mathrm{d}s$$ Let us first show that $$h(a) =0$$. This is trivial. Then let us show that $$h(b)=0$$ : $$h(b)=\int_a^b f(s)-c \, \mathrm{d}s=\int_a^b f(s) \, \mathrm{d}s-\left . cs \right |_{s=a}^b=\int_a^bf(s)ds-(b-a)c=\\ \int_a^bf(s)ds-\int_a^bf(s)ds=0$$

Now let us show that $$\int_a^b (f(x)-c)^2 dx =0$$:

$$\int_a^b (f(x)-c)^2 dx =\int_a^b (f(x)-c)h'(x) dx =\\ \int_a^b f(x)h'(x) dx + \int_a^b -ch'(x) dx = \\ \int_a^b f(x)h'(x) dx -c (h(b)-h(a))$$

But we know from our assumption $$\int_a^b f(x)h'(x) \, \mathrm{d}x=0$$, so the first term vanishes and as $$h(a)=h(b)=0$$ so does the second term. This results in $$\int_a^b (f(x)-c)^2 dx =0$$

As the integral of a nonnegative function is zero then the function $$(f(x)-c)^2$$ must be zero a.e as well, we have $$f(x)=c \qquad x\in[a,b]$$

• Assuming, as pointed out in the comments over at the question, that the $b$ in $h(x)=\int_a^b f(x)-c \, \mathrm{d}x$ has to be an $x$; this seems like the approach they want me to take. However, isn't this just one example of how $\int_a^b f(x)h'(x) \, \mathrm{d}x = 0 \Rightarrow f(x)=c$? How is this supposed to be an exhaustive proof of the lemma? – JoieNL Feb 16 '19 at 12:02
• Exactly. I think it might have been just a typo. Could you express your reasoning why it is not an exhaustive proof ? – Maksim Feb 16 '19 at 12:56
• What you're doing here is proving that $\int_a^b f(x)h'(x) \, \mathrm{d}x=0 \Rightarrow f(x)=c$ for one specific $h(x)$, right? The lemma suggests that this must be true for all $h$ that meet the specified conditions, so I presume it must be proven for all $h$, not just for this one. – JoieNL Feb 16 '19 at 13:25
• Well, the lemma assumes you have a $f$ which fulfills the condition of the integral equality for all $h$. This is the assumption, and has not to be proven. We just have to prove that $f=c$. – Maksim Feb 16 '19 at 14:11
• Oh wait of course, you're absolutely right. I'm sorry for the confusion. I'll make sure to accept your answer once there is some clarity on the supposed typo. – JoieNL Feb 16 '19 at 16:51

First of all, assuming that $$f$$ has only a finite number of discontinuities, and noting that the integral is unaffected if we delete these points from $$[a,b],$$ we may assume without loss of generality that $$f$$ is continuous on all of $$[a,b]$$.

The second fact we will use is the following: if $$f>0$$ is continuous on $$[a,b]$$, and if $$\int^b_a f(x)dx=0,$$ then $$f=0$$ on $$[a,b].$$ To prove this, suppose it is false. Then there is an $$x_0\in [a,b]$$ such that $$f(x_0)=\delta >0,$$ and now the continuity of $$f$$ gives us a neighborhood $$a such that $$f(x)<\delta/2$$ on $$(a',b')$$. This implies that $$\int^b_a f(x)dx\ge \int^{b'}_{a'}f(x)dx\ge \frac{\delta (b'-a')}{2}>0$$ which is a contradiction.

Now, if we define $$h(x)=\int^x_a (f(s)-c)ds$$ where $$c$$ is as in the hint, then $$h'(x)=f(x)-c$$ and a direct substitution shows that $$\int^b_a (f(x)-c)^2dx=0,$$ and so by what we just proved, $$(f(x)-c)^2=0$$ and thus $$f(x)=c$$.