IM Gelfand and SV Fomin, Calculus of Variations
Sec. 3, Lemma 2: Reconstruction and Walkthrough

Mark Vuletic

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Source: IM Gelfand, SV Fomin. 1963. Calculus of Variations. Mineola, NY: Dover. Tr. RA Silverman. p. 10-11.

Problem: Sec. 3, Lemma 2 states:

If $$\alpha(x)$$ is continuous in $$[a,b]$$, and if $$\int^b_a \alpha(x)h'(x) dx = 0$$ for every function $$h(x) \in D_1(a,b)$$ such that $$h(a) = h(b) = 0$$, then $$\alpha(x) = c$$ for all $$x$$ in $$[a, b]$$, where $$c$$ is a constant. (p. 10)

In the lemma, $$D_1(a,b)$$ is the space of all functions defined on $$(a,b)$$ with continuous first derivatives. Being a mere mortal, and a non-Russian one at that, I found Gelfand and Fomin's proof difficult to understand at first, so I reconstructed it in a way I found easier to follow.

Reconstruction and walkthrough: The key point of strategy Gelfand and Fomin use to prove this lemma is to show that given the conditions specified by the lemma, there is a constant $$c$$ such that $$\int^b_a [\alpha(x) - c]^2 dx = 0$$. From this, it follows that $$\alpha(x) - c = 0$$ and therefore that $$\alpha(x) = c$$.

Here is how I reconstruct the proof. We start with four stipulations, some of which I phrase differently than Gelfand and Fomin:1,2

$$\alpha(x) \textsf{ is continuous in } [a,b] \tag{1}$$

$$\forall f \in D_1(a,b) [f(a) = f(b) = 0] \rightarrow \int^b_a \alpha(x) f'(x) = 0 ] \tag{2}$$

$$c =_{\textsf{df}}\frac{1}{b-a} \int^b_a \alpha(x) dx \tag{3}$$

$$h(x) =_{\textsf{df}} \int^x_a [ \alpha(\xi) - c] d\xi \tag{4}$$

From 1, 4, and the Fundamental Theorem of Calculus, we get

$$h(x) \in D_1(a,b) \tag{5}$$

We note that 4 makes $$h(a)$$ an integral from $$a$$ to $$a$$, and hence equal to zero. Given my choice to write out $$c$$, working through $$h(b)$$ requires a bit of legwork,3 but when we are done we have

$$h(a) = h(b) = 0 \tag{6}$$

From 4 and the Fundamental Theorem of Calculus we have

$$h'(x) = \alpha(x) - c \tag{7}$$

Some algebra and basic calculus4 gives us

$$\int^b_a [\alpha(x) - c]^2 dx = \int^b_a [\alpha(x)h'(x)]dx - c[h(b) - h(a)] \tag{8}$$

From 2, 5, and 6 it immediately follows that

$$\int^b_a [\alpha(x)h'(x)]dx = 0 \tag{9}$$

6 implies

$$c[h(b) - h(a)] = 0 \tag{10}$$

9 and 10 jointly imply

$$\int^b_a [\alpha(x)h'(x)]dx - c[h(b) - h(a)] = 0 \tag{11}$$

Substitute 11 into 8 to get

$$\int^b_a [\alpha(x) - c]^2 dx = 0 \tag{12}$$

From the fact that squares of real-valued functions are positive semidefinite, we know that 12 implies

$$\alpha(x) - c = 0 \tag{13}$$

And a stroke of amazing insight allows us to move from 13 to

$$\alpha(x) = c \tag{14}$$

And there you go.

## Notes

1 In 2, I have substituted $$f$$ for the $$h$$ in the original rendition of the lemma. The change is purely notational, but something I needed for personal clarity. Although $$h$$, as it occurs in the lemma, is bound by a quantifier, and thus a variable that ranges over functions, Gelfand and Fomin reuse $$h$$ in their proof (see step 4 in my reconstruction) to represent a specific function. This is one of several practices that is common among mathematicians, but irritating and confusing to someone like me, who learned formal logic before higher mathematics. If my notation confuses you, then just substitute $$h$$ back to $$f$$ wherever you see $$f$$.

2 In 3, Gelfand and Fomin define $$c$$ indirectly, but I write it out.

3 This where just leaving $$c$$ indirectly defined, in Gelfand and Fomin's style, really becomes an advantage, since it makes it immediately evident that $$h(b) = 0$$. However, I like to see (when possible) all of the moving pieces that are hidden in indirect definitions, so if you want to join me in showing $$h(b) = 0$$ the messier way, here are the details: $$h(b) = \int^b_a [\alpha(\xi) - c)]d\xi \text{ [From 4]}$$ $$= \int^b_a \alpha(\xi)d\xi - \int^b_a c d\xi$$ $$= \int^b_a \alpha(\xi)d\xi - \int^b_a \left[\frac{1}{b-a}\int^b_a \alpha(x)dx \right]d\xi \text{ [Using 3]}$$ $$= \int^b_a \alpha(\xi)d\xi - \frac{1}{b-a}\left[\int^b_a \alpha(x)dx \right]\xi\Bigg|^{\xi=b}_{\xi=a}$$ $$= \int^b_a \alpha(\xi)d\xi - \frac{1}{b-a}\left[\int^b_a \alpha(x)dx \right](b-a)$$ $$= \int^b_a \alpha(\xi)d\xi - \int^b_a \alpha(x)dx$$ $$= 0$$

4 Details: $$\int^b_a [\alpha(x) - c]^2 dx = \int^b_a [\alpha(x) - c][\alpha(x) - c]$$ $$= \int^b_a [\alpha(x) - c]h'(x)dx \text{ [Using 7]}$$ $$= \int^b_a [\alpha(x)h'(x) - ch'(x)]dx$$ $$= \int^b_a [\alpha(x)h'(x)]dx - c \int^b_a h'(x)dx$$ $$= \int^b_a [\alpha(x)h'(x)]dx - c[h(b) - h(a)]$$

Last updated: 25 November 2016

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