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I'm confused by the highlighted remark in the text quoted below. Isn't $df_x$ linear by the very definition? At least Rudin and Apostol define derivative to be a linear map.

Actually, as I write this question I perhaps came up with an answer, but since I'm not sure whether I'm right, let me state my version here and see whether it is correct.

I think what is meant in the text below is that their definition of the derivative at $x$ evaluated at $h$ does not assume linearity. Whereas the definitions of Rudin and Apostol assume linearity, and then they (at least Apostol) prove (see Theorem 12.3 here Expressing directional derivative in terms of partial derivatives) that the derivative of $f$ at $x$ evaluated at $h$, $df_x(h)$, actually equals the directional derivative of $f$ at $x$ in the direction of $h$. Am I interpreting what is going on correctly?

Also, if we assume the definition given in the text below, how to prove that $df_x$ is linear? I don't see how it follows after playing with the quotient inside the $\lim$ sign. (I don't have Spivak at hand.)

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  • $\begingroup$ I think you have to require that $df_x$ is linear (but that might be true only for infinite-dimensional vector spaces), there exists examples in which the directional derivative exists at all points but $df_x$ is not linear. Anyway, if $f$ is differentiable, meaning that $f(x+h)=f(x)+df_x(h) + o(|h|)$, then $df_x$ is automatically linear. Please disregard this comment if you feel it confuses you, I am writing in a hurry. $\endgroup$
    – Giuseppe Negro
    Jan 21, 2018 at 22:22
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    $\begingroup$ What they define in this text is the Gâteaux derivative, which is not necessarily linear (but for smooth functions it is, and coincides with the Fréchet derivative). Rudin and Apostol define the Fréchet derivative, which is linear. See here for some relations between the two. $\endgroup$
    – Daniel Fischer
    Jan 21, 2018 at 22:33

1 Answer 1

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You can show that it is linear if the function is at least $C^1$.

You have $$df_x(ah)=\lim_{t\rightarrow 0}{{f(x+tah)-f(x)}\over t}=a \lim_{t\rightarrow 0}{{f(x+tah)-f(x)}\over{at}}.$$ Write $u=at$, then you have $$a\lim_{t\rightarrow 0}{{f(x+tah)-f(x)}\over{at}}=a \lim_{u\rightarrow 0}{{f(x+uh)-f(x)}\over u}=adf_x(h).$$

$$df_x(h_1+h_2)=\lim_{t\rightarrow 0}{{f(x+t(h_1+h_2))-f(x)}\over t}=\lim_{t\rightarrow 0}{{f(x+t(h_1+h_2))-f(x+th_1)}\over t}+ \lim_{t\rightarrow 0}{{f(x+th_1)-f(x)}\over t}.$$

Suppose that the function is $C^1$, you have $f(x+th_1+th_2)=f(x+th_1)+tdf_{x+th_1}(h_2)+O(uh_2)$ you deduce that $f(x+t(h_1+h_2))-f(x+th_1)=tdf_{x+th_1}(h_2)+O(th_2)$ This implies that $\lim_{t\rightarrow 0}{{f(x+t(h_1+h_2))-f(x+th_1)}\over t}=\lim_{t\rightarrow 0}df_{x+th_1}(h_2)=df_x(h_2)$.

Using the expression above, you deduce that $df_x(h_1+h_2)=df_x(h_1)+df_x(h_2)$.

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