Exercise 2.2.11

Question

Let $f: \mathbb{R}^{2} ightarrow \mathbb{R}$ be the function defined by $f(x, y):=\frac{x y}{x^{2}+y^{2}}$ when $(x, y) 
eq(0,0)$, and $f(x, y)=0$ otherwise. Show that for each fixed $x \in \mathbb{R}$, the function $y \mapsto f(x, y)$ is continuous on $\mathbb{R}$, and that for each fixed $y \in \mathbb{R}$, the function $x \mapsto f(x, y)$ is continuous on $\mathbb{R}$, but that the function $f: \mathbb{R}^{2} ightarrow \mathbb{R}$ is not continuous on $\mathbb{R}$.

Prakash Anand · Accepted Answer

\begin{proof}
 From the properties of continuous functions (sum, product, quotient), it is clear that for each fixed $x \in \mathbb{R}$, the function $y \mapsto f(x, y)$ is continuous on $\mathbb{R}$, and that for each fixed $y \in \mathbb{R}$, the function $x \mapsto f(x, y)$ is continuous on $\mathbb{R}$. Note that continuity could be a little delicate around $(0,0)$. But if $x=0$ then $f(0, y)=0$, and if $y=0$ then $f(x, 0)=0$, so continuity is $\mathrm{OK}$ around $(0,0)$.\par

To show that $f$ is not continuous let us look at what happens around the point $(0,0)$ with the path $x=y$. Consider $\lim _{x ightarrow 0} f(x, x)=\lim _{x ightarrow 0} \frac{x^{2}}{x^{2}+x^{2}}=\lim _{x ightarrow 0} \frac{1}{2}=\frac{1}{2} 
eq f(0,0)=0 .$ Hence $f$ is not continuous.
\end{proof}

Exercise 2.2.11

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Exercise 2.2.11

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