# A question involving Frechet differentiability

Let $X, Y$ be real normed spaces and $U \subset X$ open subset. In "Nonlinear functional analysis and applications" edited by Louis B. Rall, we have the followint definition (page 115)

A map $F : U \to Y$ is said to be $\textbf{Frechet differentiable}$ at $x_0 \in U$ if there exists a continuous linear operator $L(x_0) : X \to Y$ such that the following representation holds for every $h \in X$ with $x_0 + h \in U$, $$F(x_0 + h) - F(x_0) = L(x_0) h + r(x_0; h)$$ where $$\underset{h \to 0}{lim} \frac{\| r(x_0; h) \|}{\| h \|} =0.$$

How can I calculate $r(x_0; h)$?

For example, at page 118 (Example 1.6) we have the following function: $$f(x_1, x_2) = \frac{x_1^3 x_2}{x_1^4 + x_2^2} \; \; \text{if} \; \; (x_1, x_2) \neq (0, 0) \;\; \text{and} \;\; f(x_1, x_2) = 0, \; \; \text{if} \; \; (x_1, x_2) = (0, 0)$$ and $$r(0; h) = \frac{h_1^3 h_2}{h_1^4 + h_2^2} \;\; \text{if} \;\; h \neq 0.$$ Why does $r(0; h)$ have this form?

Thank you!

• You meant $r(h_1,h_2)$? – zhw. Apr 20 '15 at 19:26
• No. $r(0; h)$. It says in the book. I think that it is a notation. – g.pomegranate Apr 20 '15 at 19:30
• In fact, we have $r(x_0; h)$ and we consider the Frechet differentiability in $x_0$. – g.pomegranate Apr 20 '15 at 19:36

## 2 Answers

Your first line is indeed a definition of $r$: \begin{equation*} r(x_0; h) = F(x_0 + h) - F(x_0) - L(x_0) \, h. \end{equation*}

As mentioned by lalala, the typical road for proving Fréchet differentiability is the following:

• Find a candidate for $L(x_0)$. This can be done by formal calculations or by considering directional derivatives.
• Define the remainder $r$ by the above line and prove $r(x_0;h) = o(\lVert h \rVert)$.

First you have to find a candidate for $L(0,0)$, given by the partial derivatives. This gives you the $0$ matrix. Then observe that $f(0,0)=0$. So $r(0;h)$ is equal to $f(h1,h2)$.