# Confusion about linearity of Gateaux Derivative

I have the following formulation of the Gateaux derivative for functions $f:\mathbb{R}^m\to\mathbb{R}^n$. Let $f(x) = \sum_{i=1}^n f_i(x)e_i$ where $e_i$ forms a basis for $\mathbb{R}^n$, and $f_i:\mathbb{R}^m\to\mathbb{R}$. Then \begin{aligned} \text{d}f(u;\psi) & = \lim_{t\to 0}\frac{f(u+t\psi)-f(u)}{t} \\ & = \sum_{i=1}^n\left(\lim_{t\to0}\frac{f_i(u+t\psi)-f_i(u)}{t}\right)e_i \\ & = \sum_{i=1}^n(\nabla f_i\cdot\psi)e_i \end{aligned} From this it is easy to see that $\text{d}f(u;\psi)$ is linear in $\psi$. Furthermore, it would seem that the existence of the Gateaux derivative is equivalent to the existence of the partial derivatives of $f_i$, which is also what P. Drabek and J. Milota state in their book "Methods of Nonlinear Analysis" (p. 118). However, it appears that not only linearity is "false", but equivalence of existence is false as well. Consider $m=2$ and $n=1$, and $$f(x,\ y) = \frac{x^3}{x^2+y^2}$$ with $f(0,\ 0) = 0$. The partial derivatives do not exist at $0,\ 0$, as the limit is path dependent, but the Gateaux derivative does exist.

This, I can mentally justify. The partial derivatives do exist along a certain path (say, $(x, y)\to(0,\ 0)$ as $(at, bt)\to(0,\ 0)$), so that's rectified. However, as the Wikipedia article for Gateaux derivatives states, $$\text{d}f(0,0;a,b) = \frac{a^3}{a^2+b^2}$$ (and $0$ at $a=b=0$). This function however is not linear. So basically my question is: what is happening? Part of my confusion must lie in the fact that it's 3am where I live, but I've thought about this for a while and I just can't seem to rectify it in my head. Why am I getting this conflict?

• because the gateaux derivative of $f$ is linear at $v$ if and only if $f$ is differentiable at $v$.
– Surb
Apr 5, 2017 at 7:37

You say that

$$df(u,\psi)=\sum_{i=1}^n(\color{red}{\nabla f_i}\cdot\psi)e_i$$ From this it is easy to see that $\text{d}f(u;\psi)$ is linear in $\psi$.

and

Consider $m=2$ and $n=1$, and $$f(x,\ y) = \frac{x^3}{x^2+y^2}$$ with $f(0,\ 0) = 0$. The $\color{red}{\text{partial derivatives do not exist at }(0,0)}$, as the limit is path dependent, but the Gateaux derivative does exist.

Do you see where is the problem?

• I must be going in circles. In my head, I know existence of partial derivatives is equivalent to existence of Gateaux derivatives, which implies the Gateaux derivative is linear in the second argument. However these partial derivatives don't exist, but the Gateaux derivative exists. That equivalence is confusing me. Is Gateaux differentiability equivalent to existence of the partial derivatives only when the Gateaux derivative linear? Apr 5, 2017 at 7:44
• No... "existence of partial derivatives is equivalent to existence of Gateaux derivatives" this is wrong. Just look at your example. Differentiable map implies Gateaux differentiable (and the directional derivative is linear). Gateaux differentiable do not implies differentiable.
– Surb
Apr 5, 2017 at 7:48
• The Gateaux derivative is not necessarily linear when it exists.
– Surb
Apr 5, 2017 at 7:50
• Then P. Drabek and J. Milota are simply wrong. They say, quote, "It is easy to see that [the Gateaux derivative] exists if and only if [the $i$th partial derivative] exists for all $i$." Unless "if and only if" meant something different back in 2007. Apr 5, 2017 at 7:51
• Well I don't know these guys, but the sentence "It is easy to see that the Gateaux derivative exists if and only if the iith partial derivative exists for all ii" is wrong. However, the sentence: "It is easy to see that the [Gateaux derivative exists and is linear] if and only if the [$i$th partial derivative exists for all $i$]" is correct.
– Surb
Apr 5, 2017 at 7:53