Differential of Differentiable Functional is Unique/Lemma

Lemma
If $\phi[y;h]$ is a linear functional w.r.t. h and

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

Then $\phi[y;h]=0$.

Proof
The proof will be obtained by reaching a contradiction.

Suppose a linear functional satisfying $\phi[y;h_0]\ne 0$ for some $h_0\ne 0$.

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

Now, define $h_n=\frac{h_0}{n}$ and $m=\frac{\phi[y;h_0]}{\left\vert{h_0}\right\vert}$. 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 $$\lim_{n\to\infty}\frac{\phi[y;h_n]}{\left\vert{h_n}\right\vert}=\lim_{\left\vert{h_n}\right\vert\to0}\frac{\phi[y;h_n]}{\left\vert{h_n}\right\vert}=0.$$

However, using the linearity of $\phi[y;h_0]$ w.r.t. $h_0$, $$\lim_{n\to\infty}\frac{\phi[y;h_n]}{\left\vert{h_n}\right\vert}=\lim_{n\to\infty}\frac{\phi[y;\frac{h_0}{n}]}{\left\vert{\frac{h_0}{n}}\right\vert}=\lim_{n\to\infty}\frac{n~\phi[y;h_0]}{n~\left\vert{h_0}\right\vert}=\lim_{n\to\infty}\frac{~\phi[y;h_0]}{\left\vert{h_0}\right\vert}=\frac{\phi[y;h_0]}{\left\vert{h_0}\right\vert}=m\ne0.$$

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


 * : $\S 1.3$: The Variation of a Functional. A Necessary Condition for an Extremum