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Find an example for a $f\colon X\to Y$ which is Gâteaux-differentiable in a point $x_0$ but not continuous in this point $x_0$.

I am not good in finding examples but I thought of $$ f\colon\mathbb{R}\to\mathbb{R}, f(x):=\begin{cases}0, & x\leq 1\\1, & x>1\end{cases} $$ I guess on the one hand this function is not continuous in $x_0=1$ but the Gâteaux derivative $$ \eta=Df(1)[h]=\lim\limits_{t\to 0}\frac{f(1+th)-f(1)}{t}=0 $$ exists to my opinion.

Is this an appropriate example?


Isn't the function $$ F\colon\mathbb{R}\to\mathbb{R}, F(x):=\begin{cases}1, & x\neq z\\0, & x=z\end{cases} $$ Gâteaux-differentiable in $z$ and not continuous in $z$?

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Somebody should probably say it: No, the function in the Edit is not Gâteaux-differentiable at $z$. Proof: Use the definition. – Did Feb 8 '13 at 23:14
up vote 2 down vote accepted

The function $f:\mathbb R\to\mathbb R$ defined by $f(x)=\mathbf 1_{x\gt1}$ is not Gâteaux-differentiable at $x=1$ since $t^{-1}(f(x+th)-f(x))=t^{-1}$ diverges when $t\to0$, $t\gt0$, for $x=1$ and $h=1$. In fact, in dimension $1$, Gâteaux-differentiability implies continuity.

On the other hand, consider a point $z$ in $\mathbb R^2$, a circle $\bar C$ passing through $z$, $C=\bar C\setminus\{z\}$, and the function $g=\mathbf 1_C$.

For every $h\ne(0,0)$ in $\mathbb R^2$, the line $z+h\mathbb R$ does not meet $C$ in a neighborhood of $z$, that is, there exists some positive $t_{h,z}$ such that $\{z+th\mid|t|\lt t_{h,z}\}\cap C$ is empty. (To see this, note that $C$ intersects each line passing through $z$ at one point at most, which is never $z$.)

In particular, for every $|t|\lt t_{h,z}$, $t^{-1}(g(z+th)-g(z))=0$ hence the Gâteaux-differential of $g$ at $z$ in the direction $h$ exists and is $0$, although $g$ is not continuous at $z$ since the point $z$ is in the boundary of $C$.

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Could you please try to explain me how one can find such an example? I always have big problems with constructing fitting examples: For example here: How can one start to constrct such an example - what is the procedure of finding such an example? I do not have much fantasy so maybe it would help me to hear how one can find logically an example like the one given by you. – math12 Jan 30 '13 at 21:06
Yes: spiral. $ $ – Did Jan 30 '13 at 21:12
Sorry, what do you mean? – math12 Jan 30 '13 at 21:14

Let $f : \mathbb{R}^2 \to \mathbb{R}$ be defined by $\displaystyle f : (x,y) \mapsto \left\{ \begin{matrix} (0,0) & \text{if} \ (x,y)=(0,0) \\ \frac{x^4y}{x^6+y^3} & \text{otherwise} \end{matrix} \right.$. Then $f$ is Gâteaux differentiable at $(0,0)$ with $D(f)(0,0) \equiv 0$, but $f$ is not continuous at $(0,0)$, since $f(x,x^2)= \frac{1}{2}$.

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