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.

  • $\begingroup$ Should the definition of $h(x)$ have an $x$ in it somewhere? $\endgroup$ – Greg Martin Feb 15 '19 at 18:10
  • $\begingroup$ Right, if the definition were $h(x) = \int_a^x [f(s) - c] ds$, then this all follows through. $\endgroup$ – Tom Chen Feb 15 '19 at 19:07
  • $\begingroup$ 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. $\endgroup$ – JoieNL Feb 15 '19 at 21:30
  • $\begingroup$ 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. $\endgroup$ – 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 $f(x)=c \qquad x\in[a,b]$

  • $\begingroup$ 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? $\endgroup$ – JoieNL Feb 16 '19 at 12:02
  • $\begingroup$ Exactly. I think it might have been just a typo. Could you express your reasoning why it is not an exhaustive proof ? $\endgroup$ – Maksim Feb 16 '19 at 12:56
  • $\begingroup$ 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. $\endgroup$ – JoieNL Feb 16 '19 at 13:25
  • $\begingroup$ 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 $f=c$. $\endgroup$ – Maksim Feb 16 '19 at 14:11
  • $\begingroup$ 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. $\endgroup$ – 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 $a<a'<x_0<b'<b$ 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 $f(x)=c$.


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