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If $f$ is differentiable at $x$, the map $h\mapsto f(x+h)-f(x)$ should be approximately linear. The scalar multiplicativity can be seen by noting that $$\lim_{h\to 0}\frac{f(x+ch)-f(x)}{h} = \lim_{h\to 0}c\frac{f(x+ch)-f(x)}{ch}= c\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\quad \text{and}\\ \lim_{t\to 0}\frac{f(x+ctv)-f(x)}{t}=\lim_{t\to 0}c\frac{f(x+ctv)-f(x)}{ct}=c\lim_{t\to 0}\frac{f(x+tv)-f(x)}{t},$$

where on the second line, $x$ and $v$ are vectors. The additivity of this map is less easy to see for me. How can we show it? That is, how to show that $$\lim_{t\to 0}\frac{f(x+t(v+w))-f(x)}{t} = \lim_{h\to 0}\frac{f(x+tv)-f(x)}{t}+\lim_{h\to 0}\frac{f(x+tw)-f(x)}{t}?$$

Eric Auld
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  • Are you assuming differentiability? If so, you need only show that $Df(v)$ is the directional derivative of $f$ in the direction $v$, which is not too hard. If you are not assuming differentiability, I do not think you can get anywhere. In particular, a function can be Gâteaux differentiable with its Gâteaux derivative nonlinear.

    http://en.wikipedia.org/wiki/G%C3%A2teaux_derivative

     – Steven Gubkin Feb 28 '14 at 22:35
  • Could you help me understand what you're doing? In the title you ask about showing that the directional derivative is linear but then in the question you don't actually use the expression for the directional derivative (which would include the direction vector). – Rudy the Reindeer Oct 20 '14 at 22:54
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    @MattN. $h$ is the direction vector in the first part, $v$ in the second part. It's confusing because in the article you link to, $h$ is a scalar. :) – Eric Auld Oct 22 '14 at 00:58

1 Answers1

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The subtlety here, Eric, is that when you're working with a function $f\colon\Bbb R^n\to\Bbb R$, the definition of differentiability is far stronger than the assumption of existence of all directional derivatives. So you must use linearity as part of the definition; you cannot derive it (pun intended). (In particular, consider $$f(x,y) = \begin{cases} \frac{x^2y}{x^2+y^2}\,, & (x,y)\ne (0,0) \\ 0\,, & (x,y)=(0,0)\end{cases}.$$ This function has all directional derivatives at $(0,0)$ but is not differentiable at $(0,0)$.) You need linearity of the derivative to prove directional derivatives are linear.

A good question to pose is this: Suppose you have linearity in the directional derivatives. Is this sufficient to guarantee differentiability? (Hint: There are functions discontinuous at $(0,0)$ with all directional derivatives at $(0,0)$ equal to $0$ !!)

Ted Shifrin
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