Gateaux derivative

Remarks

For $X = \R^n$ and $Y = \R$, $df(x; h)$ is the directional derivative of $f$ along $h$. Some authors do not ask $h \mapsto df(x; h)$ to be linear or continuous. The definition works without change for locally convex spaces. If $f$ has a Fréchet derivative $Df(x)$, then $f$ is Gateaux differentiable at $x$ with $df(x; h) = Df(x) h$.

Example: let $H$ be a real Hilbert space and $f(x) = \norm{x}^2$. Then $f(x + t h) = \norm{x}^2 + 2t \ip{x}{h} + t^2 \norm{h}^2$, so $df(x; h) = 2\ip{x}{h}$, which is continuous and linear in $h$.

Non-example: on $\R^2$, let $f(x, y) = x^3 / (x^2 + y^2)$ and $f(0, 0) = 0$. For $h = (a, b) \neq 0$, $f(t a, t b) / t = a^3 / (a^2 + b^2)$, so $df(0; h)$ exists for every $h$. But $h \mapsto a^3 / (a^2 + b^2)$ is not linear, so $f$ is not Gateaux differentiable at $0$.

A Gateaux differentiable map need not be Fréchet differentiable. On $\R^2$, let $g(x, y) = 1$ when $y = x^2$ and $(x, y) \neq 0$, and $g = 0$ elsewhere. Each line through $0$ meets the parabola in at most one point other than $0$, so $dg(0; h) = 0$ for all $h$. But $g$ is not continuous at $0$.