# Differential of Differentiable Functional is Unique

## Proof

### Lemma

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

Let:

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

Then :

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

$\Box$

Let $J \sqbrk y$ be a differentiable functional.

Suppose the differential of $J \sqbrk y$ is not uniquely defined.

Then at least $2$ different forms of this exist:

$\Delta J \sqbrk {y; h} = \phi_1 \sqbrk {y; h} + \epsilon_1 \size h$
$\Delta J \sqbrk {y; h} = \phi_2 \sqbrk {y; h} + \epsilon_2 \size h$

where:

$\phi_1 \sqbrk {y; h}$ and $\phi_2 \sqbrk {y; h}$ are linear functionals
$\epsilon_1, \epsilon_2 \to 0$ as $\size h \to 0$.

Because $\size h \to 0$, the whole expression $\to 0$ faster than $\size h$ due to constraints on $\epsilon$.

Hence $\epsilon_1 \size h$, $\epsilon_2 \size h$ and $\paren {\epsilon_1 - \epsilon_2} \size h$ are all infinitesimals of order higher than $1$ relative to $\size h$.

Consequently, subtraction of one from the other leads to:

$(1): \quad \phi_1 \sqbrk {y; h} - \phi_2 \sqbrk {y; h} = \paren {\epsilon_2 - \epsilon_1} \size h$

Therefore, $\phi_1 \sqbrk {y; h} - \phi_2 \sqbrk {y; h}$ is an infinitesimal of order higher than $1$ relative to $\size h$.

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

Rearranging the terms of $(1)$:

$\dfrac {\phi_1 \sqbrk {y; h} - \phi_2 \sqbrk {y; h} } {\size h} = \epsilon_2 - \epsilon_1$

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

$\displaystyle \lim_{\size h \mathop \to 0} \frac {\phi_1 \sqbrk {y; h} - \phi_2 \sqbrk {y; h} } {\size h} = 0$

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

Hence:

$\phi_1 \sqbrk {y; h} - \phi_2 \sqbrk {y; h} = 0$

That is, there is only one form the differential of a differentiable functional can take.

$\blacksquare$