# Differential of Differentiable Functional is Unique/Lemma

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## Lemma

Let $\phi \sqbrk {y; h}$ be a linear functional with respect to $h$.

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.

Aiming for a contradiction, suppose 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 $\phi \sqbrk {y; h_0}$ with respect to $h_0$:

\(\ds \lim_{n \mathop \to \infty} \frac {\phi \sqbrk {y; h_n} } {\size {h_n} }\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \frac {\phi \sqbrk {y; \frac {h_0} n} } {\size {\frac {h_0} n} }\) | Definition of $h_n$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \frac {n \, \phi \sqbrk {y; h_0} } {n \, \size {h_0} }\) | extract of $n$ through linearity | |||||||||||

\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \frac {\phi \sqbrk {y; h_0} } {\size {h_0} }\) | cancel $n$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \frac {\phi \sqbrk {y; h_0} } {\size {h_0} }\) | the limit does not depend on $n$ | |||||||||||

\(\ds \) | \(=\) | \(\ds m\) | definition of $m$ |

However, by hypothesis:

- $m \ne 0$

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

$\blacksquare$

## Sources

- 1963: I.M. Gelfand and S.V. Fomin:
*Calculus of Variations*: $\S 1.3$: The Variation of a Functional. A Necessary Condition for an Extremum