In Serge Lang's Differential and Riemannian manifold, he defined derivative between two topological vector spaces with the concept tangent to $0$:
Let $\mathbf{E}, \mathbf{F}$ be two topological vector spaces, and $\varphi$ a mapping of a neighborhood of $0$ in $\mathbf{E}$ into $\mathbf{F}$. We say that $\varphi$ is tangent to 0 if, given a neighborhood $W$ of 0 in $\mathbf{F}$, there exists a neighborhood $V$ of 0 in $\mathbf{E}$ such that $$ \varphi(t V) \subset o(t) W $$ for some function $o(t)$. If both $\mathbf{E}, \mathbf{F}$ are normed, then this amounts to the usual condition $$ |\varphi(x)| \leqq|x| \psi(x) $$ with $\lim \psi(x)=0$ as $|x| \rightarrow 0$. Let $\mathbf{E}, \mathbf{F}$ be two topological vector spaces and $U$ open in $\mathbf{E}$. Let $f: U \rightarrow \mathbf{F}$ be a continuous map. We shall say that $f$ is differentiable at a point $x_0 \in U$ if there exists a continuous linear map $\lambda$ of $\mathbf{E}$ into $\mathbf{F}$ such that, if we let $$ f\left(x_0+y\right)=f\left(x_0\right)+\lambda y+\varphi(y) $$ for small $y$, then $\varphi$ is tangent to 0 . It then follows trivially that $\lambda$ is uniquely determined, and we say that it is the derivative of $f$ at $x_0$. We denote the derivative by $D f\left(x_0\right)$ or $f^{\prime}\left(x_0\right)$. It is an element of $L(\mathbf{E}, \mathbf{F})$
I think I might understand the intuition of this tangent to 0 concept. However, why is this definition equivalent to the definition stated in the normed space case? I don't really see how to write a proof to show these definitions are equivalent.
$\|\varphi(x)\|_{F}\leq \|x\|_{E} \psi(x)$ is the definition I know before, but I'm having trouble to relate to the general definition, where it specifically refers to "given a neighborhood $W$". So why are these two definitions equivalent?
To show that the definition in the topological vector space case implies the definition in the normed space case, let $\psi :E\to [0, \infty )$ be a function such that $\lim_{x\to 0}\phi (x) = 0$ and $\| \varphi (x) \|_{F} \leq \psi (x) \|x\|_{E}$ for all $x\in E$. I am also going to assume that $\psi (0) = 0$. Define $\Phi :[0, \infty ) \to [0, \infty )$ by
$$\Phi (t) := \min \left\{ \left( \sup_{\|x\|_{E} \leq t} \psi (x)\right) , 1 \right\}.$$
Note that $\Phi$ is increasing. To show that $\lim_{t\to 0}\Phi (t) = 0$, let $\alpha \in (0,1)$. By the definition of $\psi$, there is some $\delta\in (0, \infty )$ such that if $x\in E$ and $\|x\|_{E} \leq \delta$, then $\psi (x) \leq \frac{\alpha}{2}$. By the definition of $\Phi$, there is some $x\in E$ with $0< \|x\|_{E} \leq \delta$ such that $\frac{1}{2}\Phi (\delta ) \leq \psi (x)$, from which it follows that $\Phi (\delta ) \leq \alpha$. So if $t\in [0, \delta]$, it follows that $\Phi (t) \leq \alpha$. This shows that $\lim_{t\to 0}\Phi (t) = 0$.
Let $V$ be an open neighbourhood of $0$ in $F$. If $B_{F} = \{x\in F : \|x\|_{F} < 1\}$ is the open unit ball in $F$, then there exists some $\varepsilon \in (0, \infty )$ such that $\varepsilon B_{F} \subseteq V$. As $\lim_{t\to 0}\Phi (t) = 0$, there is some $\delta \in (0, \infty )$ such that $\Phi (\delta ) < \varepsilon$. Define $U := \delta B_{E}$, where $B_{E} = \{x\in E : \|x\|_{E} < 1\}$ is the open unit ball in $E$. Define $\phi :[0, \infty ) \to [0, \infty )$ by
$$\phi (t) := \frac{2t\delta}{\varepsilon}\Phi (t\delta ).$$
It will be shown that $\phi$ has the desired properties. If $t\in (0, \infty)$, then
$$\frac{\phi (t)}{t} = \frac{2\delta}{\varepsilon}\Phi (t\delta ) .$$
It follows that $\lim_{t\to 0}\Phi (t\delta ) = 0$ and consequently that $\lim_{t\to 0}\frac{\phi (t)}{t} = 0$.
To show that $\varphi (tU) \subseteq \phi (t)V$ for all $t\in (0, \infty )$, let $x\in U$ and $t\in (0, \infty )$. If $\varphi (tx) = 0$, the result is obvious. Otherwise, suppose that $\varphi (tx) \neq 0$. Then $\varphi (tx) \neq 0$ implies that $\phi (t) > 0$, and consequently
$$\| \varphi (tx) \|_{F} \leq \psi (tx) \|tx\|_{E} \leq \Phi (t\delta )t\delta = \left( \frac{2t\delta }{\varepsilon} \Phi (t\delta ) \right) \cdot \frac{\varepsilon}{2} = \phi (t) \cdot \frac{\varepsilon}{2} < \phi (t)\varepsilon .$$
This shows that $\varphi (tx) \in \phi (t)V$. This proves the first direction.
To show that the definition in the normed space case implies the definition in the topological vector space case, note that there exists some $\delta \in (0, \infty )$ and some $\phi :[0, \infty )\to [0, \infty )$ with $\lim_{t\to 0} \frac{\phi (t)}{t} = 0$, where I will again assume that $\phi (0) = 0$, such that for all $x\in E$ and $t\in (0, \infty )$ such that $\|x\|_{E} < \delta$ and $\varphi (tx) \neq 0$, it follows that $\| \varphi (tx) \|_{F} < \phi (t)$. Note that this is equivalent to the property that $\varphi (t\delta B_{E}) \subseteq \phi (t)B_{F}$ for all $t\in (0, \infty )$, and this property follows from the definition of $\varphi$ being tangent to $0$ in the topological vector space case. Define $\psi :E \to [0, \infty )$ by
$$\psi (x) := \begin{cases} \frac{2}{\delta \|x\|_{E}}\phi (\|x\|_{E}) & \text{ if } x\in E\setminus \{0\}, \\ 0 & \text{ if } x\in \{0\}. \end{cases}$$
To show that $\lim_{x\to 0}\psi (x) = 0$, let $\varepsilon \in (0, \infty )$. As $\lim_{t\to 0}\frac{\phi (t)}{t} = 0$, there is some $r\in (0, \infty )$ such that if $t\in [0, r)$, then $\phi (t) \leq \varepsilon t$. Consequently, if $x\in E$ with $\|x\|_{E} < r$, then $\psi (x) \leq \frac{2}{\delta}\varepsilon$. It follows that $\lim_{x\to 0}\psi (x) = 0$.
It will now be shown that $\| \varphi (x) \|_{F} \leq \psi (x) \|x\|_{E}$ for all $x\in E$. Let $x\in E$. If $\varphi (x) = 0$, the result is obvious. Otherwise, suppose $\varphi (x) \neq 0$. Then there are $\lambda \in (0, \infty )$ and $y\in E$ such that $x = \lambda y$ where $\|y\|_{E} = \frac{\delta}{2}$ and $\lambda = \frac{2}{\delta}\|x\|_{E}$. As $\|y\|_{E} < \delta$, it follows that
$$\| \varphi (x) \|_{F} \leq \phi (\lambda ) = \left( \frac{2}{\delta}\cdot \frac{\phi (\lambda )}{\lambda } \right) \cdot \frac{\delta\lambda}{2} = \psi (x) \|x\|_{E}.$$
This proves the second direction.