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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 is 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:

$$ \int^b_a [\alpha(x) – c]^2 dx = \int^b_a [\alpha(x) – c][\alpha(x) – c] $$

$$ = \int^b_a [\alpha(x) – c]h'(x)dx \text{ [Using 7]}$$

$$ = \int^b_a [\alpha(x)h'(x) – ch'(x)]dx $$

$$ = \int^b_a [\alpha(x)h'(x)]dx – c \int^b_a h'(x)dx $$

$$ = \int^b_a [\alpha(x)h'(x)]dx – c[h(b) – h(a)] $$