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Assume there is a multivariable function $f:\mathbb{R}^n\to\mathbb{R}$. I want to know the relation between Gateaux differentiable and differentiable. That is, if $f$ is differentiable, is it Gateaux differentiable? If $f$ is Gateaux differentiable, is it differentiable? Thank you.

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It would be helpful if you add the definitions of both Gateaux differentiable and differentiable (since they are always precisely the same). –  Dirk May 5 '13 at 8:03
    
differentiable is what we meet in undergraduate calculus course (see, e.g., Def. 9.11 in P212 of baby Rudin). Gateaux differentiable (at x) is defined as follows: (1)the directional derivative $f'(x;v)$ exists for all $v\in\mathbb{R}^n$, (2)there exists (unique) $f_G'(x)\in\mathbb{R}^*$ such that $f'(x;v)=<f_G'(x),v>$ for all $v\in\mathbb{R}^n$, where $\mathbb{R}^*$ means the dual space of $\mathbb{R}^n$. –  Zhou Heng May 6 '13 at 0:05

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  1. $f$ is differentiable at a point $a$ if there is a vector $v$ such that $$\lim_{x\to a} \frac{|f(x)-f(a)-\langle v,x-a\rangle |}{|x-a|}=0 \tag1$$ This is also called Fréchet differentiability, to distinguish it from other kinds.

  2. $f$ is Gâteaux differentiable at a point $a$ for any nonzero vector $u$ the limit
    $$\lim_{h\to 0} \frac{ f(a+hu)-f(a) }{h} \tag2$$ exists. This is also called the directional derivative.

  3. $f$ is Gâteaux differentiable at a point $a$ with linear differential if there is a vector $v$ such that for any nonzero vector $u$ we have
    $$\lim_{h\to 0} \frac{ f(a+hu)-f(a) }{h}=\langle v,u\rangle \tag2$$ The vector $v$ is called the Gâteaux differential of $f$. Normally, the words Gâteaux differential are used only in this case, when we have linearity.

It is true that (1) implies (3). The easiest way to prove this is to rewrite (1) as $f(x)=f(a)+\langle v,x-a\rangle +r(x)$ where the remainder $r(x)$ satisfies $r(x)/|x-a|\to 0$ as $x\to a$. Then plug into (2).

It is obvious that (3) implies (2).

In spaces of dimension $n\ge 2$, (2) does not imply (3), and (3) does not imply (1). Examples are given in wikipedia article.

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Thank you!_____ –  Zhou Heng May 5 '13 at 23:55

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