Differential of Differentiable Functional is Unique/Lemma

Lemma
Let $\phi \sqbrk {y; h}$ be a linear functional $h$.

Let:
 * $\ds \lim_{\size h \mathop \to 0} \frac {\phi \sqbrk {y; h} } {\size h} = 0$

Then:
 * $\phi \sqbrk {y; h} = 0$

Proof
This will be a Proof by Contradiction.

there exists a linear functional satisfying $\phi \sqbrk {y; h_0} \ne 0$ for some $h_0 \ne 0$.

Also suppose:
 * $\ds \lim_{\size {h_0} \mathop \to 0} \frac{\phi \sqbrk {y; h_0} } {\size {h_0} } = 0$

Now, define:
 * $h_n = \dfrac {h_0} n$

and:
 * $m = \dfrac {\phi \sqbrk {y; h_0} } {\size {h_0} }$

Notice that $\size {h_n} \to 0$ as $n \to \infty$.

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


 * $\ds \lim_{n \mathop \to \infty} \frac {\phi \sqbrk {y; h_n} } {\size {h_n} } = \lim_{\size {h_n} \mathop \to 0} \frac {\sqbrk {y; h_n} } {\size {h_n} } = 0$

However, using the linearity of $\phi \sqbrk {y; h_0}$ $h_0$:

However, :
 * $m \ne 0$

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