Differential of Differentiable Functional is Unique

From ProofWiki
Jump to: navigation, search


The differential of a differentiable functional is unique.



Let $\phi \left[{y; h}\right]$ be a linear functional w.r.t. $h$.


$\displaystyle \lim_{\left \vert{h}\right\vert \to 0} \frac {\phi \left[{y; h}\right]} {\left\vert{h}\right\vert} = 0$

Then :

$\phi \left[{y; h}\right] = 0$

Let $J[y]$ be a differentiable functional.

Suppose the differential of $J[y]$ is not uniquely defined.

Then at least 2 different forms of this exist:

$\Delta J[y;h]=\phi_1[y;h]+\epsilon_1 \left\vert{h}\right\vert$

$\Delta J[y;h]=\phi_2[y;h]+\epsilon_2\left\vert{h}\right\vert$

where $\phi_1[y;h]$ and $\phi_2[y;h]$ are linear functionals, and $\epsilon_1,\epsilon_2\to 0$ as $\left\vert{h}\right\vert\to 0$.

Note that $\epsilon_1\left\vert{h}\right\vert$, $\epsilon_2\left\vert{h}\right\vert$ and $\left(\epsilon_1-\epsilon_2\right)\left\vert{h}\right\vert$ are all infinitesimals of order higher than 1 relative to $\left\vert{h}\right\vert$,

since as $\left\vert{h}\right\vert\to 0$, the whole expression $\to 0$ faster than $\left\vert{h}\right\vert$ due to constraints on $\epsilon$.

Consequently, subtraction of one from the other leads to

$\phi_1[y;h]-\phi_2[y;h]=\left(\epsilon_2-\epsilon_1\right) \left\vert{h}\right\vert$.

Therefore, $\phi_1[y;h]-\phi_2[y;h]$ is an infinitesimal of order higher than 1 relative to $\left\vert{h}\right\vert$.

Since each of the members are linear functionals, the whole term keeps the same property.

By rearranging terms of the last equation we get


Taking the limit and recalling the constraint on both $\epsilon_1$ and $\epsilon_2$ results in

$\displaystyle\lim_{\left\vert{h}\right\vert\to 0}\frac{\phi_1[y;h]-\phi_2[y;h]}{\left\vert{h}\right\vert}=0$

The given limit with the linearity of the term in the numerator allows us to use the lemma.

This concludes to $\phi_1[y;h]-\phi_2[y;h]=0$ meaning that there is only one form the differential of a differentiable functional can obtain.