# Uniqueness of the derivative in locally convex topological vector space

I need a hint of proof of uniqueness of the derivative in locally convex topological vector space (it's asserted in Lang's "Introduction to differentiable manifolds").

Define derivative of a function $f: E \to F$ between two topological vector spaces at the point $x_0$ as a linear operator $f'(x_0) \in L(E, F)$ such that ${}^2\!\!f(y) := f(x_0 + y) - f(x_0) - f'(x_0) y$ is tangent to zero (a function $\varphi$ is tangent to zero if for any neighborhood of zero $W \subset F$ there exists a neighborhood of zero $V \subset E$ such that $\varphi(tV) = o(t)W$).

Now suppose that two operators $A_1$ and $A_2$ satisfy the condition of the derviative. I need to prove then that $A_1 = A_2$. It is easy to see that there must be a neighborhood of zero $V \subset E$ such that $(A_1 - A_2)V = 0$, so we have $\operatorname{span}V \subset \ker (A_1 - A_2) \subset E$, so it is sufficient to prove that $\operatorname{span}V = E$. In case of Banach space it is very easy (any non-empty open set must contain a ball), but in a more general case it seems that this attempt fails (consider a discrete space), and I can't see another strategy.

Is the derivative in fact unique or is some condition stronger than what is stated required for it to be unique?

• If $A_1 \neq A_2$ then there exists $y$ such that $A_1 y \neq A_2 y$. Apply Hahn-Banach. – t.b. Jul 16 '11 at 12:19
• By the way: a discrete space is not a topological vector space! (Hint: consider the scalar action) Also, by definition of local convexity, a neighborhood of zero is absorbing. – t.b. Jul 16 '11 at 12:26
• Ok, so we have the scalar action $s: \mathbb{K} \times V \to V$, and $U \subset V$. Then $\pi_\mathbb{K}(s^{-1}(U)) = \mathbb{K}$, seems perfectly continuous to me. In more detail, for any $x \in V$ $s^{-1}(x) = \mathbb{K} \times \mathbb{K}x$, and this set is open, so the preimage of any set is a union of such sets, and so it's open, am I right? – Alexei Averchenko Jul 16 '11 at 12:41
• No it isn't. If $\lambda_n \to \lambda \neq 0$ and $\lambda_n \neq \lambda$ then by continuity $\lambda_n v \to v$ and this can't be if $v \neq 0$ (we're in a discrete space and $\lambda_n v \neq \lambda v$). – t.b. Jul 16 '11 at 12:44
• Did you mean $\lambda_n v \to \lambda v$? Why can't it be? I'm puzzled: why do we even need to consider limits here when the continuity is painfully obvious from the usual topological definition? – Alexei Averchenko Jul 16 '11 at 12:50

I suggested to use Hahn-Banach. Let $A = A_1 - A_2 \neq 0$. Take $z = Ay = (A_1 - A_2)y \neq 0$ (assuming that $A_1 \neq A_2$). Then by Hahn-Banach we find a continuous linear functional $\phi: F \to \mathbb{K}$ such that $\phi(z) = 1$. I let you finish this alternative proof, using that $\phi \circ A$ is continuous (I suggested this proof before reading your question entirely, there's not much difference).