Why does "t" in the formula for gateaux differentiation represent a scalar value?

Question

$$\lim \limits_{t \to 0} \frac{f(x_0+tv)-f(x_0)}{t}$$ This is the formula I have in my textbook for a function defined in $\mathbb{R}^p$ with values in $\mathbb{R}^q$ that calculates the directional derivative. I don't understand where $t$ comes from. Since it's in place of the $x_0-x$, I would assume $t = x_0-x$, but since the function takes as an argument a $p$-dimensional vector, that would mean we are trying to do vector division, which is not possible. Can someone explain what it is that I'm not understanding?

Answer 1

The direction is specified by the vector $v$ and the amount in that direction by the scalar $t$. Think, for fixed $v$, that $x_0+tv$ traces a line passing through $x_0$ in the direction $v$, as $t\in\mathbb R$. In a sense, study of the differentiability of $f:\mathbb R^p\to\mathbb R^q$ is reduced to study of the composition $t\mapsto f(x_0+tv)$, a function from $\mathbb R \to \mathbb R^q$.

The value of the derivative of this composite function depends on $v$, and if you replace $v$ with $2v$ the derivative of $f(x_0+t2v)$ with respect to $t$ at $t=0$ will of course be double that of $f(x_0+tv)$. This perhaps explains part of why the derivative of the original $f$ is usually said to be a linear transformation (or matrix, if you will), called $df(x_0)$ (or $M$ or $Df_{x_0}$ etc: textbooks differ), so that $$ \left.\frac d {dt}\right|_{t=0} f(x_0+tv) = (df(x_0))\,v$$ This recipe solves the scaling problem: if you double $v$, you will double the value of $df(x_0)v$ too.

You are right, one does not know how to divide by vectors in the obvious attempt to generalize the one-dimensional definition of the derivative. One can consider the business of looking at $tv$ with scalar $t$ and vector $v$ a kind of hack compromise between what one really wants but doesn't know how to do and what is easy to work with but is perhaps not instantly intuitive.