Necessary Condition for Existence of Functional Extrema

Question

If $J$ is a differentiable functional and has an extremum at $y_{0}$, it must be that its differential is 0. That is, $\delta J[y_{0};h] = 0$ for all admissible $h$.

We go by contradiction. Suppose for some $h_{0}$ that $\delta J[y_{0};h_{0}] \neq 0$. By definition we have that $$\lim_{||h||\to 0} \frac{\Delta J[y_{0};h] - \delta J[y_{0};h]}{||h||} = 0 .$$ So for $\varepsilon > 0$ we have for some $0 < \eta < \varepsilon$ that if $||h||<\eta$ then $$ \delta J[y_{0};h] - \varepsilon||h|| < \Delta J[y_{0};h] < \delta J[y_{0};h] + \varepsilon||h|| .$$ Let $\alpha > 0$ be small enough so that $||\alpha h_{0}|| < \eta$. Put $g = \alpha h_{0}$. We arrive at the inequalities \begin{align} \delta J[y_{0},g] - \varepsilon||g|| &< \Delta J[y_{0},g] < \delta J[y_{0},g]+\varepsilon||g|| \\ -\delta J[y_{0},g] - \varepsilon||g|| &< \Delta J[y_{0},-g] < -\delta J[y_{0},g]+\varepsilon||g||. \end{align} We see it cannot be that $\delta J[y_{0},g] \geq \varepsilon||g||$.

I can't bring this home and I'm not sure my "translation" of their definition of differentiable is correct.

Answer 1

If $J$ is a differentiable functional and has an extremum at $y_{0}$, it must be that its differential is 0. That is, $\delta J[y_{0};h] = 0$ for all $h$.

By definition we have that $$\lim_{||h||\to 0} \frac{\Delta J[y_{0};h] - \delta J[y_{0};h]}{||h||} = 0 .$$ So for $\varepsilon > 0$ we have for some $0 < \eta < \varepsilon$ that if $||h||<\eta$ then \begin{align*} \delta J[y_{0};h] - \varepsilon||h|| < \Delta J[y_{0};h] < \delta J[y_{0};h] + \varepsilon||h|| \\ -\delta J[y_{0};h] - \varepsilon||h|| < \Delta J[y_{0};-h] < -\delta J[y_{0};h] + \varepsilon||h|| \end{align*} Because $J$ is extremized at $y_{0}$, we can also make $\eta$ small enough so that $\Delta J[y_{0};h]$ is either always positive or always negative for all $||h|| < \eta$. In this case, we see it must be that $|\delta J[y_{0};h]| < \varepsilon||h||$ because otherwise $\Delta J[y_{0};h]$ and $\Delta J[y_{0};-h]$ have opposite signs. But then $$\frac{|\delta J[y_{0};h] }{||h||} < \varepsilon $$ and we see that $$\lim_{||h||\to 0}\frac{\delta J[y_{0};h]}{||h||} = 0 .$$ The following lemma shows it must be that $\delta J[y_{0};h]$ is identically 0.

If $J$ is a linear functional such that $$\lim_{||h||\to0}\frac{J[h]}{||h||} = 0 $$ then $J$ is identically 0.

Suppose for some $h_{0}$ that $\phi[h_{0}] \neq 0$. Put $$h_{n} = \frac{h_{0}}{n} \neq 0.$$ As $h_{0}$ must be bounded, we see that $\lim_{n\to\infty}||h_{n}|| = 0$ and $$\lim_{n\to \infty} \frac{\phi[h_{n}]}{||h_{n}||} = 0$$ by hypothesis. On the other hand, $$\lim_{n\to \infty} \frac{\phi[h_{n}]}{||h_{n}||} = \lim_{n\to\infty} \frac{n\phi[h_{0}]}{n||h_{0}||} = \frac{\phi[h_{0}]}{||h_{0}||} \neq 0 $$ a contradiction.

Answer 2

Say, $y$ is a local minimum of $J$. Then there exist $\epsilon > 0$ and a function $R$, defined on the $\epsilon$-ball $B_\epsilon$ around zero, such that for each $h\in B_\epsilon$ you have $$ 0\le J[y+h] - J[y] = \delta J[y,h] + R[h], $$ where $|R[h]|/\|h\|\to 0$ as $h\to 0$. This also holds for $-h$ instead of $h$, so also $-\delta J[y,h] + R[-h]\ge 0$. This gives $$ -|R[h]|\le-R[h]\le\delta J[y,h]\le R[-h]\le |R[-h]| $$ for all $h\in B_\epsilon$. In particular, $|\delta J[y,h]|\le\max\{|R[h]|,|R[-h]|\}$.

Now, fix some $h_0$ with $\|h_0\|=1$. Then for $\alpha\in (0,\epsilon)$ you have $$ |\alpha||\delta J[y,h_0]| = |\delta J[y,\alpha h_0]|\,\le\,\max\{|R[\alpha h_0]|,|R[-\alpha h_0]|\}, $$ and hence (as $|\alpha| = \|\alpha h_0\| = \|-\alpha h_0\|$), $$ |\delta J[y,h_0]|\,\le\,\max\left\{\frac{|R[\alpha h_0]|}{\|\alpha h_0\|},\frac{|R[-\alpha h_0]|}{\|-\alpha h_0\|}\right\}. $$ This holds for any $\alpha\in(0,\epsilon)$. Letting $\alpha\to 0$ yields $\delta J[y,h_0] = 0$.