Why is it that for $x^n + y^n = 1$ and $n$ is an even number, does as $n$ goes to infinity the graph approaches a square?

Question

Why is it that for $x^n+y^n= 1$ and $n$ is an even number, does as $n$ goes to infinity the graph approaches a square?

Answer 1

David gave a reasonable intuition for this but I thought a proof would be nice (and I think it is a pretty good answer to "why?"). The $p$-norm of a vector $x = (x, y) \in \mathbb{R}^2$ is defined as $\| x\|_p = (x^p + y^p)^{\frac{1}{p}}$. What we want to do is prove that the graph of $x^p + y^p = 1$ as $p$ goes to infinity approaches a square, i.e, the closed ball $B_{1}(0)$ under the maximum norm (equivalently, the set of points $(x, y) \in \mathbb{R}^2$ such that $\max\{ |x|, |y| \} = 1$). Denoting $x = (x_1, x_2)$ and $\|x\|_{\infty}=\max\{ |x_1|, |x_2| \}$ , notice that $$ \left|x_{k}\right|=\left(\left|x_{k}\right|^{p}\right)^{\frac{1}{p}} \leq\left(\sum_{j=1}^{2}\left|x_{j}\right|^{p}\right)^{\frac{1}{p}}=\|x\|_{p} $$

and therefore $\|x \|_{\infty} \leq \|x\|_p$. In particular, $$ \|x\|_{\infty} \leq \liminf _{p \rightarrow \infty}\|x\|_{p} $$ On the other hand, since $ \left|x_{j}\right| \leq\|x\|_{\infty} \forall j \in \mathbb{N} $, we have for any $q < p$: $$ \|x\|_{p}=\left(\sum_{j=1}^{2}\left|x_{j}\right|^{p-q} \cdot\left|x_{j}\right|^{q}\right)^{\frac{1}{p}} \leq\|x\|_{\infty}^{\frac{p-q}{p}} \cdot\left(\sum_{j=1}^{2}\left|x_{j}\right|^{q}\right)^{\frac{1}{p}}=\|x\|_{\infty}^{1-\frac{q}{p}} \cdot\|x\|_{q}^{\frac{q}{p}} $$ So we get:

$$ \limsup _{p \rightarrow \infty}\|x\|_{p} \leq \limsup _{p \rightarrow \infty}\left(\|x\|_{\infty}^{1-\frac{q}{p}} \cdot\|x\|_{q}^{\frac{q}{p}}\right)=\|x\|_{\infty} \cdot 1 $$ and therefore $$ \limsup _{p \rightarrow \infty}\|x\|_{p} \leq\|x\|_{\infty} \leq \liminf _{p \rightarrow \infty}\|x\|_{p} $$ which proves $\lim _{p \rightarrow \infty}\|x\|_{p}$ exists and equals $\|x\|_{\infty}$, as desired (evidently your question follows immediately from this result).

Clearly this proof can be generalized. But I think an even more interesting question is why the graph of $x^p + y^p = 1$ behaves the way it does when $p$ is odd.

Answer 2

Because for any number $x$ less than $1$, $\lim\limits_{n \to \infty}x^n \to 0$. It is only at $x =1$ that $\lim\limits_{n \to \infty} x^n = 1.$