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Let $f:\mathbb{R}^N\rightarrow\mathbb{R}^M$ be a function which is Gâteaux differentiable and let $J_f\in\mathbb{R}^{M\times N}$ be its Jacobian matrix.

Is it true that the Gâteaux derivative of $f$ along a direction $v\in\mathbb{R}^N$ is equal to the matrix-vector product $J_f \cdot v$?

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  • $\begingroup$ I will upvote this later today when my upvote limit refreshes $\endgroup$ Jul 21, 2016 at 16:41

2 Answers 2

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As you probably already know, this is true if $f$ is Frechet differentiable.

However, there is no reason why this should be true if $f$ is only Gateaux differentiable but not Frechet differentiable, because then the directional derivatives need not be linearly related to each other, i.e. as is stated in Wikipedia, "the Gateaux differential of a function may be nonlinear".

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

The Jacobian matrix specifies the directional derivatives in the directions of the standard basis elements (as you probably already know, these are also called partial derivatives). But if $f$ is Gateaux but not Frechet differentiable, then the directional derivative in, for example $v=e_1+e_2$ may not equal $\partial_1 f + \partial_2 f$, which means that we could not use the Jacobian matrix to compute it.

The main idea here being that matrix multiplication represents composition with a linear function, so that if the Gateaux derivative is a nonlinear function of the directional derivatives, then it cannot be represented by a matrix.

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The answer in general is no. The classic counter example is given by the function

$$ f(x,y)=\left\{ \begin{array}{cc} \frac{x^2y}{x^4+y^2} & (x,y)\neq (0,0),\\ 0 & (x,y)=(0,0) \end{array} \right. $$

You can check that at $(0,0)$ this function has all the directional derivatives. However, it is neither differentiable, nor continuous. In particular, the partial derivatives are both zero, but along all other directions its derivative is nonzero.

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