Is this a typo about directional derivative?

Question

I'm reading about the directional derivative:

Let $(E,\|\cdot\|)$ and $(F,\|\cdot\|)$ be Banach spaces over the field $\mathbb{K}$, and $X$ an open subset of $E$. A function $f: X \rightarrow F$ is differentiable at $a \in X$ if there is $A \in \mathcal{L}(E, F)$ such that $$f(x)=f\left(a\right)+A\left(x-a\right)+o\left(\left\|x-a\right\|\right) \quad\left(x \rightarrow a\right)$$

The author continues to define the directional derivative:

and prove a proposition:

We define $g:\mathbb K \rightarrow F, \quad t \mapsto f\left(x_{0}+t v\right)$. Our goal is to find the derivative of $g$ at $t=0$.

Because $f$ is differentiable at $x_0$, $$ f(x)=f\left(x_{0}\right)+\partial f\left(x_{0}\right)\left(x-x_{0}\right) + o(\left\|x-x_{0}\right\|) \quad (x \to x_0)$$ for all $x \in X$. It follows that $$\begin{aligned}g(t) &= f(x_0+tv) \\ &=f\left(x_{0}\right)+\partial f\left(x_{0}\right)\left((x_0+tv)-x_{0}\right)+o(\left\|(x_0+tv)-x_{0}\right\|) \quad ((x_0+tv) \to x_0)\\ &= g\left(0 \right)+\partial f\left(x_{0}\right)\left(tv\right)+o(\left\|tv\right\|) \quad (t \to 0) \\&= g\left(0\right)+ \partial f\left(x_{0}\right)\left(v\right) \cdot (t-0) + o(\left\|t\right\|) \quad (t \to 0) \end{aligned}$$

Let $\partial g(0) \in \mathcal{L}(\mathbb K, F)$ be the derivative of $g$ at $t=0$. It follows that $$g(t)= g(0) + \partial g(0)(t-0) + o(\left\|t\right\|) \quad (t \to 0)$$

To sum up, we have $$\begin{aligned}g(t) &=g\left(0\right)+ \partial f\left(x_{0}\right)\left(v\right) \cdot (t-0) + o(\left\|t\right\|) \quad (t \to 0)\\ &= g(0) + \partial g(0)(t-0) + o(\left\|t\right\|) \quad (t \to 0)\end{aligned}$$

Hence $\partial f\left(x_{0}\right)\left(v\right) \cdot (t-0) = \partial g(0)(t-0)$ and thus $\partial f\left(x_{0}\right)\left(v\right) \cdot t = \partial g(0)(t)$. Because $(t)$ on the right hand is just the input of the function $\partial g(0)$, we have $\partial f\left(x_{0}\right)\left(v\right) \cdot t = \partial g(0)$.

In my understanding, $\partial g(0)(t)$ denote the value of the function $\partial g(0)$ at $t$. On the contrary, $\partial f\left(x_{0}\right)\left(v\right) \cdot t$ denoted the product of $\partial f\left(x_{0}\right)\left(v\right)$ and $t$.

My question:

I could not understand why the answer given in my textbook is just $\partial f\left(x_{0}\right)\left(v\right)$, which is lack of $t$.

Could you please elaborate on this point?

Answer 1

Now, I see what confuses you. It's the fact that $\mathcal L(\mathbb K,F) \cong \mathbb K$. The isomorphism is $\phi : \mathcal L(\mathbb K,F)\to\mathbb K$, defined by $\phi(T) = T(1)$. So you can identify $T\in\mathcal L(\mathbb K,F)$ with $T(1)\in\mathbb K$.

You have $t\partial f(x_0)(v) = \partial g(0)(t)$. This is correct. Now, note that $t = t\cdot 1$ and that $\partial g(0)$ is linear. So, $\partial g(0)(t) = t\partial g(1)$. Division by $t$ gives $\partial f(x_0)(v) = \partial g(0)(1) = \partial g(0)$ in the sense of identification.

But you're right, they are not the same when you're rigorous.

Answer 2

I think the identification referred to in the comments can be explained by simply looking at a differentiable function $f:\mathbb R\to \mathbb R.$

In elementary calculus, if $f$ is differentiable at $x_0$, then there is a number $f'(x_0)$ and a function $r:\mathbb R\to \mathbb R$ such that $f(x+h)=f(x_0)+f'(x_0)h+r(h)$ and $r(h)/h\to 0$ as $h\to 0.$

Using the more general formulation, if $f$ is differentiable at $x_0$, then there is a linear transformation $f'(x_0):\mathbb R\to \mathbb R$ and a function $r:\mathbb R\to \mathbb R$ such that $f(x+h)=f(x_0)+f'(x_0)h+r(h)$ and $r(h)/h\to 0$ as $h\to 0.$

The defining equation is $\textit{the same}$ in both cases. Only the perspective changes. This is because of the isomorpshism $\mathbb R\cong \mathcal L(\mathbb R,\mathbb R)$ that sends the number $a$ to the linear transformation $T$, that sends $h$ to $ah$. Then, $T(1)=a$, which is exactly the relation between the $\textit {linear transformation}\ T:=f'(x_0)$ and the $\textit{number}\ f'(x_0)$

In your case, we have $g:\mathbb K \rightarrow F, \quad t \mapsto f\left(x_{0}+t v\right)$ which is a composition $g=f\circ \phi$ where $\phi(t)=x_0+tv.$ The derivative at $t=0$ is a linear transformation $g'(0):\mathbb K\to F$ that satisfies

$g(x+h)=g(x_0)+f'(x_0)h+r(h)$ and $r(h)/h\to 0$ as $h\to 0$ and the same identification applies, because $F\cong \mathcal L(\mathbb K,F):\ f'(x_0)$ is an element of $F$ and also a $T\in L(\mathbb K,F)$ via the identifiaction $T(1)=f'(x_0).$