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Find which values of $\alpha, \beta > 0$ make $f: \mathbb{R^2} \rightarrow \mathbb{R}$ continuous.

$$ \left\{ \begin{array}{c} f(x,y)=\frac{|{x}|^{\alpha}|{y}|^{\beta}}{x^2 + y^2}, {\ \ \ }(x,y) \neq (0,0) \\ 0, {\ \ \ }(x, y)=(0,0) \end{array} \right. $$

The limit of $f$ in $(0,0)$ has to be 0 for $f$ to be continuous. Let's consider the sequence $\{x_k\} =\{ (\frac{1}{k},\frac{1}{k})\}$.

It converges to $(0,0)$ and its points are all different from $(0,0)$, so if $f$ is continuous in $(0,0)$ then $\{f(x_k)\}$ must converge to 0.

$\{f(x_k)\}=\frac{|{\frac{1}{k}}|^{\alpha}|{\frac{1}{k}}|^{\beta}}{(\frac{1}{k})^2 + (\frac{1}{k})^2}=\frac{{\frac{1}{{k}^{\alpha}}}{\frac{1}{{k}^{\beta}}}}{\frac{2}{k^2}}=\frac{k^2}{2{k}^{\alpha + \beta}};$

So if $\alpha + \beta =2$ then $\{f(x_k)\}$ converges to $\frac{1}{2}$, and if $\alpha + \beta <2$ then $\{f(x_k)\}$ does not converge. We have then that if $\alpha+\beta \le 2$ $f$ is not continuous. What I can't prove is that if $\alpha+\beta \gt 2$ then $f$ is continuous (I don't even know if it's true).

Yagger
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1 Answers1

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use the $AM-GM$ inequality $$x^2+y^2\geq 2|xy|$$ it is $$\frac{a+b}{2}\geq \sqrt{ab}$$ for all $a,b\geq 0$$

Dr. Sonnhard Graubner
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  • Could you explain how to use it? If $x^2+y^2\geq 2|xy|$ then $\frac{|{x}|^{\alpha}|{y}|^{\beta}}{x^2 + y^2} \le \frac{|{x}|^{\alpha}|{y}|^{\beta}}{2|xy|}$ but I don't know how to continue. – Yagger Dec 06 '17 at 21:24