Differential of Differentiable Functional is Unique

Theorem
The differential of a differentiable functional is unique.

Proof
This will be a proof by contradiction.

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 $\left\vert{h}\right\vert\to 0$.

Consequently, subtraction of one from the other leads to

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

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 $$\frac{\phi_1[y;h]-\phi_2[y;h]}{\left\vert{h}\right\vert}=\epsilon_2-\epsilon_1.$$ Taking the limit and recalling the constraint on both $\epsilon_1$ and $\epsilon_2$ results in $$\lim_{\left\vert{h}\right\vert\to 0}\frac{\phi_1[y;h]-\phi_2[y;h]}{\left\vert{h}\right\vert}=0.$$ 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.