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I have seen the proof done different ways, but none using the norm definitions provided.

Given: $||x||_p = (|x_1|^p+|x_2|^p)^{1/p}$ and $||x||_\infty = max(|x_1|,|x_2|)$

Prove: $\lim_{p\rightarrow\infty}\|x\|_p = \|x\|_\infty$

I have looked at the similar questions: The $ l^{\infty} $-norm is equal to the limit of the $ l^{p} $-norms. and Limit of $\|x\|_p$ as $p\rightarrow\infty$ but they both seem to use quite different approaches (we have not covered homogeneity so that is out of the question, and the other uses a different definition for the infity norm).

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user3784030
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    Read the first question again, it is exactly what you are looking for. – Quang Hoang Aug 29 '14 at 05:26
  • Perhaps what is confusing you is that those questions are looking at the $p$-norm and $\infty$-norm on the space of sequences instead of on $\mathbb{R}^2$. It's not hard to ungeneralize the math done there. – JimmyK4542 Aug 29 '14 at 05:29

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Hint: WLOG, assume $|x_1| \ge |x_2|$. Then, $\left(|x_1|^p+|x_2|^p\right)^{1/p} = |x_1|\left(1+\left(\dfrac{|x_2|}{|x_1|}\right)^p\right)^{1/p}$.

Can you show that this approaches $|x_1|$ as $p \to \infty$?

JimmyK4542
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