# Question regarding the definition of differentiability

In multivariable calculus, we say $f: U \subseteq \mathbb{R}^d \to \mathbb{R}^m$ is differentiable at $x_0 \in U \subseteq \mathbb{R}^d$, $U$ open if there exists a linear map $T: \mathbb{R}^d \to \mathbb{R}^m$ such that

$$\lim_{h \to 0 } \frac{f(x_0 + h) - f(x_0) - T(h)}{||h||} = 0$$

Equivalently, we say that $f$ is differentiable if $\exists T$ such that

$$f(x_0 + h) - f(x_0) = T(h) + o(||h||)$$ as $h \to 0$. Now, if we put $h = tx$. Then if $x \neq 0$, have that $h \to 0 \implies t \to 0^+$ assuming $t$ to be positive. Then, we have

$$f(x_0 + xt) - f(x_0) = T(xt) + o(||tx||) \implies \frac{ f(x_0 + xt) - f(x_0)}{t} = \frac{t T(x)}{t} + t \;o(||x||)$$

$$\therefore \frac{ f(x_0 + xt) - f(x_0)}{t} = T(x) + o(||x||)t$$

In the limit $t \to 0^+$, then we obtain that

$$T(x) = \lim_{t \to 0^+}\frac{ f(x_0 + xt) - f(x_0)}{t}$$

Is this a correct argument to show that $T$ can be expressed as above? thanks for any feedback.

• your argument looks good, however, beware that if you construct $T$ as you indicate then its not sufficient to show differentiability of $f$. What you write for $T$ is the directional derivative of $f$ at $x_o$ in the $x$-direction. In the case that $f$ is differentiable then certainly the differential reproduces directional derivatives. However, there are examples of functions which have directional derivatives in all directions and yet fail to be differentiable. For example: math.stackexchange.com/questions/372070/… – James S. Cook Mar 3 '14 at 20:15

Technically, this definition would essentially yield the Jacobian matrix applied in the direction of $x$.
Note that when you define $h=tx$, then you're assuming $h$ gets small along the path defined by $x$. So essentially this becomes a directional derivative. In fact, this is precisely the definition of a directional derivative.
However, the linear map must apply when $h$ gets small generally, not just along some $x$.
• So, $T(x)$ as defined in my last line does not make sense ? – ILoveMath Mar 3 '14 at 20:07
• Correct. What you're defining is the the directional derivative in the direction of $x$, but a function can have directional derivatives in every direction, but not be differentiable at a point. See the example here: en.wikipedia.org/wiki/… – Emily Mar 3 '14 at 20:09