4
$\begingroup$

It's known that real exponentiation $x^y$ is continuous in each variable, but is real exponentiation jointly continuous in both the exponent and the base?

I considering the function $(0,\infty)\times\mathbb{R}\to\mathbb{R}$ sending $(x,y)\mapsto x^y$. If $(x_n,y_n)$ is a sequence approaching $(x,y)$, then $x_n\to x$ and $y_n\to y$ individually. Since both $\exp$ and $\ln$ are continuous functions, does it suffice to just say $x^y=\exp(y\ln (x))$, so $$ \begin{align*} \lim_{n\to\infty} x_n^{y_n} &=\lim_{n\to\infty} \exp\left(y_n\ln(x_n)\right)\\ &= \exp\left(\lim_{n\to\infty}y_n\ln(x_n)\right)\\ &= \exp\left(\lim_{n\to\infty}y_n\ln\left(\lim_{n\to\infty}x_n\right)\right)\\ &= \exp\left(y\ln(x)\right)=x^y? \end{align*} $$

I am just not sure of my method since my experience is limited to single variable functions.

$\endgroup$
  • 3
    $\begingroup$ Yes, that suffices. You could also split it into steps, $(x,y) \mapsto (\log x, y)$ is continuous because $\log$ is. $(w,y) \mapsto w\cdot y$ is continuous, $\exp$ is continuous, so the composition of the three is also continuous. $\endgroup$ – Daniel Fischer Jul 8 '13 at 1:23
  • 2
    $\begingroup$ To add to the discussion: if you consider the function over the extended domain $[0,\infty)\times \mathbb R \to \mathbb R$, you'll find that the function is no longer jointly continuous. In particular, the limit $$\lim_{(x,y)\to(0,0)}x^y$$ does not exist. $\endgroup$ – Omnomnomnom Jul 8 '13 at 1:34
  • $\begingroup$ @Daniel's comment is really an answer. $\endgroup$ – Jonas Meyer Jul 8 '13 at 1:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.