# Differential of Differentiable Functional is Unique/Lemma

## 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 :

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

## Proof

This will be a proof by contradiction.

Aiming for a contradiction, suppose there exists a linear functional satisfying $\phi \left[{y; h_0}\right] \ne 0$ for some $h_0 \ne 0$.

Also suppose:

$\displaystyle \lim_{\left \vert{h_0}\right\vert \to 0} \frac{\phi \left[{y; h_0}\right]} {\left\vert{h_0}\right\vert} = 0$

Now, define:

$h_n = \dfrac {h_0} n$

and:

$m = \dfrac{\phi \left[{y; h_0}\right]} {\left\vert{h_0}\right\vert}$

Notice that $\left\vert{h_n}\right\vert \to 0$ as $n \to \infty$.

Hence, from the assumption of the limit it should hold that:

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

However, using the linearity of $\phi \left[{y; h_0}\right]$ w.r.t. $h_0$:

 $\displaystyle \lim_{n \to \infty} \frac {\phi \left[{y; h_n}\right]} {\left\vert{h_n}\right\vert}$ $=$ $\displaystyle \lim_{n \to \infty} \frac {\phi \left[{y; \frac {h_0} n}\right]} { \left\vert{\frac {h_0} n}\right\vert}$ Definition of $h_n$ $\displaystyle$ $=$ $\displaystyle \lim_{n \to \infty} \frac {n \, \phi \left[{y; h_0}\right]} {n \, \left\vert{h_0}\right\vert}$ extract of $n$ through linearity $\displaystyle$ $=$ $\displaystyle \lim_{n \to \infty} \frac {\phi \left[{y; h_0}\right]} {\left\vert{h_0}\right\vert}$ cancel $n$ $\displaystyle$ $=$ $\displaystyle \frac {\phi \left[{y; h_0}\right]} {\left\vert{h_0}\right\vert}$ the limit does not depend on $n$ $\displaystyle$ $=$ $\displaystyle m$ definition of $m$

However, by hypothesis:

$m \ne 0$

Hence, the contradiction is achieved and the initial statement of the lemma holds.

$\blacksquare$