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The following is a proof of a lemma from page 10 of Calculus of Variations by Gelfand and Fomin (tr. Silverman):

Lemma 2: If $\alpha(x)$ is continuous in $[a,b]$, and if $$\int_a^b\alpha(x)h'(x)\ dx = 0$$ for every function $h(x)$ [that is continuously differentiable on the interval $[a,b]$] such that $h(a)=h(b)=0$, then $\alpha(x)=c$ for all $x$ in $[a,b]$ where $c$ is a constant.

Proof. Let $c$ be defined by the condition $$\int_a^b[\alpha(x)-c]\ dx=0,$$ and let $$h(x)=\int_a^x[\alpha(\xi)-c]\ d\xi,$$ so that $h(x)$ automatically belongs to [the set of continuously differentiable functions on $[a,b]$] and satisfies the conditions $h(a)=h(b)=0$. Then, on the one hand, $$\int_a^b[\alpha(x)-c]h'(x)\ dx=\int_a^b\alpha(x)h'(x)\ dx-c[h(b)-h(a)]=0,$$ while on the other hand, $$\int_a^b[\alpha(x)-c]h'(x)\ dx=\int_a^b[\alpha(x)-c]^2\ dx.$$ It follows that $\alpha(x)-c=0$, i.e., $\alpha(x)=c$, for all $x$ in $[a,b]$.

This proof appears to show that $\alpha(x)=c$, and I can't find any steps that appear to be in error.

However, I seem to have found a contradiction: suppose that $\alpha(x)=c$ causes $\int_a^b\alpha(x)h'(x)\ \mathrm dx=0$. Let $\alpha_k(x)=k\alpha(x)=kc$ for some $k\in \mathbb R$. Then

$$\int_a^b\alpha_k(x)h'(x)\ \mathrm dx=\int_a^bk \alpha(x)h'(x)\ \mathrm dx=k\int_a^b \alpha(x)h'(x)\ \mathrm dx=0$$

So now the situation is this: the proof from the book clearly shows $\alpha(x)$ must be $c$ determined from $\int_a^b\alpha(x)-c\ \mathrm dx=0$, and yet, any $\alpha(x)=kc$ appears to satisfy the original condition that $\int_a^b\alpha(x)h'(x)\ \mathrm dx = 0$. Something is wrong: either the proof has an error (unlikely) or my reasoning is incorrect.

Can anyone help me understand the flaw in the argument? Thanks

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  • $\begingroup$ You say "...from $\int_a^b\alpha(x)-c\ \mathrm dx=0$, and yet, any $\alpha(x)=kc$ appears ...", but the $c$ in the integral would not be the same as the $c$ in the equality on the right. Basically, if you have $\alpha_k(x)$ then you also have $c_k$ where $c_k=kc$ $\endgroup$ Commented yesterday
  • $\begingroup$ Sorry, I'm a little lost here. Would you be willing to elaborate a little more? $\endgroup$ Commented yesterday
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    $\begingroup$ If $\alpha_k(x)=kc$ then $\int_a^b( \alpha_k(x)-c)dx = \int_a^b (kc-c)dx=c(k-1)(b-a)\neq 0$. Basically, if you take $\alpha_k(x)$ to be $kc$ then the $c^{'}$ (I've used $c^{'}$ to denote it is new) defined by $\int_a^b(\alpha_k(x)-c^{'})dx=0$ is different or $c\neq c^{'}$. $\endgroup$ Commented yesterday
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    $\begingroup$ Okay, maybe my confusion comes from not really understanding the original proof. They're trying to argue that since the original integral must hold for all choices of $h$, then we can concoct any $h$ we want to show that $\alpha$ must be a constant. But, we chose an $\alpha(x)$ in order to find $h(x)$ in order to find $\alpha(x)$.... so the argument is a little circular. $\endgroup$ Commented yesterday
  • $\begingroup$ The statement of the theorem to be proved starts with a couple of "if" clauses: if $\alpha(x)$ has the properties described, then it has yet another property. It is quite standard and not circular at all to suppose that a function has the "if" properties. What is important is that we assume only the "if" properties, and don't make any additional unproven assumptions about $\alpha.$ So we don't actually know the functions $\alpha$ or $h$ the way we know a function such as $f(x)=x^2.$ We just know some facts about $\alpha$ and $h$ that we are able to use in the proof. $\endgroup$ Commented 18 hours ago

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The statement of the theorem requires that we give an $\alpha$, then we verify that for all choices of suitable $h$ an integral is zero, from which we conclude that $\alpha$ is a constant function. The theorem does not tell you what that constant is and the construction of the theorem allows that constant to depend on whichever $\alpha$ you have at the start.

You give an example where $\alpha = c$ is a constant function and although you don't write down the result, the theorem gives that $\alpha$ is a constant function. (In fact, it's $c$.)

You give another example where $\alpha_k = kc$ is a constant function. You verify that the integral condition holds for all suitable $h$. From this you correctly conclude that $\alpha_k$ is constant. (In fact, it's $kc$.)

Note that this is true for any pair of constant functions. One is a constant multiple of the other and you can perform exactly the same "trick" using one in the theorem to show that the other is constant. What would be a problem would be using a constant function to show that a non-constant function satisfies the theorem, which you have not done.

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  • $\begingroup$ I've come back to this after a night of sleep, and now I believe I understand the reasoning. It goes like this: Suppose we had an $\alpha(x)$, which, like all functions, admits a value $c$ such that $\int_a^b\alpha(x)-c\ \mathrm dx=0$. Then, the condition imposed by the integral against $h'(x)$ forces $\alpha(x)$ to "have always been" identically equal to $c$. Is it me, or is this proof really quite tricky to follow? $\endgroup$ Commented 21 hours ago
  • $\begingroup$ @Mahkoe Not all functions allow you to define $c$ by saying $\int_a^b(\alpha(x)-c)\ \mathrm dx=0.$ But the functions that have the "if" conditions of the theorem do, and that's what you need. I looked at that formula and said to myself, isn't $c$ just the mean value of $\alpha(x)$ on $[a,b]$? Knowing that $\alpha$ is continuous on $[a,b],$ it's integrable and has a mean value. We then proceed to use that mean value as shown. $\endgroup$ Commented 18 hours ago
  • $\begingroup$ @Mahkoe : Theorem is easier to follow if, when you see $\int \alpha -c$, you think something like "$\alpha$ has a mean, and that mean is $c$". This comes more readily by filling in your web of concepts to include a broad range of particular examples of stuff. And expressing the mean in this "integral judo" way is common in an integral-oriented setting. ... $\endgroup$ Commented 11 hours ago
  • $\begingroup$ @Mahkoe : ... Maybe it's easier to think about : what function would always have the same sign as $\alpha$ minus its mean everywhere (maximizing the value of its integral)? It's $\alpha$ minus its mean. So $h' = \alpha$ minus $\alpha$'s times $\alpha$ minus its mean always nonnegative. Then accumulate from $a$ to get $h$. (Here, "accumulate from $a$" means the integral $\int_a^x h'(\xi)\,\mathrm{d}\xi$, which is the (unique) antiderivative of $h'$ guaranteed to be $0$ at $x = a$.) $\endgroup$ Commented 11 hours ago
  • $\begingroup$ @Mahkoe : When first running into these "some integral does something nice for every function in this set of functions", I would think of a standard basis for that set of functions. What you really want is do something with $\alpha$ that gives you such a function that breaks the integral relation, so you should be poking around the set of functions that are close to $\alpha$. Examples: $\alpha$, $\alpha \pm \text{ constant}$, $-\alpha$, $|\alpha{}|$, stuff like that, and you may need "project the function I thought of onto the set of test functions" to get a working test function. $\endgroup$ Commented 11 hours ago
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The key words in the proof are, "Let $c$ be defined by the condition ... ." This phrase is followed by a condition that defines $c$ in terms of $\alpha$: $$ \int_a^b[\alpha(x)-c]\,\mathrm dx=0. \tag1 $$

Assuming $b \neq a$ (see the note at the end of this answer), Equation $(1)$ is equivalent to $$ \int_a^b \alpha(x)\,\mathrm dx - \int_a^b c\,\mathrm dx = \int_a^b \alpha(x)\,\mathrm dx - (b - a) c = 0, $$ that is, $$ c = \frac1{b - a} \int_a^b \alpha(x)\,\mathrm dx. \tag2 $$

So what Equation $(1)$ says (assuming $b > a$, again see the note) is that $c$ is the mean value of $\alpha(x)$ on the interval $[a,b].$

All the proof is saying is that $\alpha(x)$ on the interval $[a,b]$ is identically equal to its own mean value on that interval. You can make that mean value be anything you like as long as $\alpha(x)$ is identically equal to that value, as you found afterward. There's no contradiction.

On the other hand, if you define a number $c$ by $$ \int_a^b[\alpha(x)-c]\,\mathrm dx=0 $$ as the proof does and then say that $\alpha(x) = kc$ identically, you've just said that $$ \int_a^b[kc-c]\,\mathrm dx=0. $$ What does that say?

If you are saying you can define a constant function whose constant value is different from its own mean value, that's a contradiction.

Note: There is a slight gap in the proof: Equation $(1)$ defines $c$ only if $b \neq a.$ If $b = a$ then the integral is zero no matter what value we give to $c.$ I suppose the idea was that $b > a$ so that $[a,b]$ is a non-empty interval of positive length, and the person writing the problem simply neglected to state that assumption.

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Put simply, the proof shows that we must have $\alpha(x)\equiv c$ for some explicit $c$ that depends on $\alpha$. Of course $\alpha(x)\equiv c'$ works for any constant $c'$, but all that means is that the $c$ defined at the start of the proof will turn out to be equal to $c'$.

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