Equivalent definition of Differentiability of function of two variable.

Question

I know the definition of Fréchet derivative of function between two normed space.and one can think it is a generalization of our usual definition of derivative of real function.But in a book i faced (equivalent) definition for real valued function of two variable.so can i have hint how can i show this definition is equivalent to Fréchet derivative : I dont even know how to start as in definition which i faced they define two error term what can i do from Frechet definition is only 1 error term.so how to get another?

Answer 1

Ok, this is the definition of the Frechet derivative for a function $f: X\to Y$ between the normed spaces $X,Y$ at the point $z$. $$f(z+h)=f(z)+G(h)+g(h),\ g(h)\in o(||h||_X)$$ Where $G(h)$ is a continuous linear operator between $X$ and $Y$. And $g(h)\in o(||h||_X)$ means that when $h\to0$, then $\ \frac{||g(h)||_Y}{||h||_X}\to 0$. In the case of a function of two variables we have $X=\mathbb{R}^2$, $Y=\mathbb{R}$, the norms are the standard euclidean norm and the absolute value, $z=(x,y)$.

So the Frechet derivative becomes $$f(x+\delta x, y+\delta y)=f(x, y)+G(\delta x, \delta y)+g(\delta x, \delta y),\ g(\delta x, \delta y)\in o(\sqrt{(\delta x)^2+(\delta y)^2})$$ also note that any linear function $G(\delta x, \delta y)$ in this case has the form $G(\delta x, \delta y)=A\delta x + B\delta y$. So the only thing to show is the equivalence between the single error term $g(\delta x, \delta y)\in o(\sqrt{(\delta x)^2+(\delta y)^2})$ and the double term $\delta x\phi(\delta x, \delta y)+\delta y\psi(\delta x, \delta y)$.

The first direction, we have the single term $g(\delta x, \delta y)\in o(\sqrt{(\delta x)^2+(\delta y)^2})$. Define also the function $$\operatorname{sign}(t)=\begin{cases}-1&,\ t<0 \\ 1&,\ t>0 \\ 0&,\ t=0\end{cases}$$ And write $g(\delta x, \delta y)$ as $$g(\delta x, \delta y)=\delta x\operatorname{sign}(\delta x)\frac{g(\delta x, \delta y)}{|\delta x|+|\delta y|}+\delta y\operatorname{sign}(\delta y)\frac{g(\delta x, \delta y)}{|\delta x|+|\delta y|}$$ So we have $$\phi(\delta x, \delta y)=\operatorname{sign}(\delta x)\frac{g(\delta x, \delta y)}{|\delta x|+|\delta y|}$$ $$\psi(\delta x, \delta y)=\operatorname{sign}(\delta y)\frac{g(\delta x, \delta y)}{|\delta x|+|\delta y|}$$ And both functions tend to $0$ as $\delta x,\delta y$ tend to zero simultaneously because $$\bigg|\operatorname{sign}(\delta x)\frac{g(\delta x, \delta y)}{|\delta x|+|\delta y|}\bigg|=\bigg|\operatorname{sign}(\delta y)\frac{g(\delta x, \delta y)}{|\delta x|+|\delta y|}\bigg|=\frac{|g(\delta x, \delta y)|}{|\delta x|+|\delta y|}\leq\frac{|g(\delta x, \delta y)|}{\sqrt{(\delta x)^2+(\delta y)^2}}\to0$$ for $(\delta x, \delta y)\to0$ due to $g(\delta x, \delta y)\in o(\sqrt{(\delta x)^2+(\delta y)^2})$.

The reverse direction. We have the two terms $\phi(\delta x, \delta y),\ \psi(\delta x, \delta y)$. Then note that $$\delta x\phi(\delta x, \delta y)+\delta y\psi(\delta x, \delta y)\in o(\sqrt{(\delta x)^2+(\delta y)^2})$$ because $$\frac{|\delta x\phi(\delta x, \delta y)+\delta y\psi(\delta x, \delta y)|}{\sqrt{(\delta x)^2+(\delta y)^2}}\leq\frac{(|\delta x|+|\delta y|)(|\phi(\delta x, \delta y)|+|\psi(\delta x, \delta y)|)}{\sqrt{(\delta x)^2+(\delta y)^2}}\leq$$ $$\leq2(|\phi(\delta x, \delta y)|+|\psi(\delta x, \delta y)|)\to0$$ for $(\delta x, \delta y)\to0$. So we name $$g(\delta x,\delta y)=\delta x\phi(\delta x, \delta y)+\delta y\psi(\delta x, \delta y)\in o(\sqrt{(\delta x)^2+(\delta y)^2})$$