---
tag: 0762WAE49P
kind: statement
role: proposition
name: Uniqueness of the derivative
msc: [26B05]
uses: [08EVNYZF42]
sources:
  - url: https://en.wikipedia.org/wiki/Fr%C3%A9chet_derivative
    locator: Definition
---

## Statement

Let $V$ and $W$ be normed vector spaces over $\mathbb{R}$, let $U \subseteq V$ be open, let $f \colon U \to W$, and let $x \in U$.
If the bounded linear maps $A$ and $B$ from $V$ to $W$ are both [[Fréchet derivatives|08EVNYZF42]] of $f$ at $x$, then $A = B$.

## Remarks

The proof uses that $U$ is open. For a unit vector $v$ and small $t > 0$, the point $x + t v$ lies in $U$, and $\|t A v - t B v\| = o(t)$. Divide by $t$ and let $t \to 0$ to get $A v = B v$.
So it is correct to speak of the derivative $Df(x)$.
On a set that is not open, for example a single point, a derivative within the set need not be unique.

## Explanations

### L1 human

Two different linear maps differ by a definite amount in some direction, at the scale of the step.
If both were good approximations, their difference would be tiny compared with the step in every direction. This forces the difference to be $0$.