This question is probably pretty stupid, but I can't figure it out so…
I'm trying to read Gelfand and Fomin's "Calculus of Variations" and on page 12, for some functional $J$ on a normed vector space $S$ they first define:
$$\Delta J [y;h] = J[y+h] – J[y], y,h\in S$$
and then say if we can write:
$$\Delta J[y;h] = \delta J[y;h]+ \epsilon ||h||$$
where $\delta J$ is a functional linear in $h$ and $\lim_{||h||\to 0} \epsilon = 0$, then we call $\delta J[y;h]$ the variation of $J$ at $y$. Quick sanity check for myself: in general, $\epsilon$ is a functional of $h$, right? Why don't we write:
$$\Delta J[y;h] = \delta J[y;h] + \epsilon[y;h] ||h||?$$
Anyway, a few pages later, in the proof of theorem 2, they assert that since $\lim_{||h||\to 0} \epsilon =0 $ then there must exist some $\delta$ s.t. $\forall h \in S$ s.t. $||h||<\delta$ we must have that the sign of $\delta J[y;h]$ is the same as the sign of $\Delta J[y;h]$. Why is this the case?
Best Answer
This has also bothered me for a while, but it can be eventually shown quite simply using an analogous argument from page 12.
Suppose $\delta J[h_0] \neq 0$ for some $h_0$ in the normed linear space where the functional is defined. We then set
$$h_n = \frac{h_0}{n}$$
so that $||h_n|| \rightarrow 0$ as $n \rightarrow \infty$. Therefore,
\begin{align} \Delta J[h_n] & = \delta J[h_n]+\epsilon||h_n|| \\ & = \frac{1}{n} ( \delta J[h_0]+\epsilon||h_0|| ) \\ \end{align}
We notice that the second term in the parenthesis tends to $0$ as $n \rightarrow \infty$, since $\epsilon \rightarrow 0$ as $n \rightarrow \infty$ (i.e. $||h_n|| \rightarrow 0$). So eventually $\Delta J[h_n]$ and $\delta J[h_0]$ will share the same sign.
However, $y$ maximises $J[y;h]$ (for example), so for sufficiently large $n$ we have $\Delta J[h_n] \geqslant 0$ and therefore, $\delta J[h_0] \geqslant 0$ and especially, $\delta J[h_0] > 0$, since $\delta J[h_0] \neq 0$.
Now let us set
$$h'_0 = -h_0$$
so that
$$\delta J[h'_0] = -\delta J[h_0] < 0$$
We then run through the same above argument with
$$h'_n = \frac{h'_0}{n}$$
and conclude once again that $J[h'_0]>0$, which leads to a contradiction.