# Is it correct to say $\text{max}(x_1,x_2,\dots,x_n)=\lim_{p \to \infty} \frac{x_1^p+x_2^p+\dots+x_n^p}{x_1^{p-1}+x_2^{p-1}+\dots+x_n^{p-1}}$

There are many possible way to represent the maximum function, I came up with this one:

$$\text{max}(x_1,x_2,\dots,x_n)=\lim_{p \to \infty} \frac{x_1^p+x_2^p+\dots+x_n^p}{x_1^{p-1}+x_2^{p-1}+\dots+x_n^{p-1}}$$

$$x_1,x_2,\dots,x_n \geq 0$$

Numerically seems to be true. I'm not sure if my proof is correct though.

Let the following condition hold (without loss of generality, except for the case when $x_1=x_k$ for some $k>1$, but I'm not going to consider it for now):

$$x_1 > x_2 \geq \dots \geq x_n$$

Now let's divide both numerator and denominator by $x_1^{p-1}$:

$$\frac{x_1^p+x_2^p+\dots+x_n^p}{x_1^{p-1}+x_2^{p-1}+\dots+x_n^{p-1}}=\frac{x_1+\frac{x_2^p}{x_1^{p-1}}+\dots+\frac{x_n^p}{x_1^{p-1}}}{1+\frac{x_2^{p-1}}{x_1^{p-1}}+\dots+\frac{x_n^{p-1}}{x_1^{p-1}}}$$

Now one part of the expression poses no trouble when taking the limit:

$$\lim_{p \to \infty} \frac{x_1}{1+\frac{x_2^{p-1}}{x_1^{p-1}}+\dots+\frac{x_n^{p-1}}{x_1^{p-1}}}=x_1$$

As for the rest of the expression, I'm not so sure:

$$\lim_{p \to \infty} \frac{\frac{x_2^p}{x_1^{p-1}}+\dots+\frac{x_n^p}{x_1^{p-1}}}{1+\frac{x_2^{p-1}}{x_1^{p-1}}+\dots+\frac{x_n^{p-1}}{x_1^{p-1}}}=\lim_{p \to \infty} \frac{x_2^p+\dots+x_n^p}{x_1^{p-1}+x_2^{p-1}+\dots+x_n^{p-1}}$$

It should be equal to $0$ if my conjecture is correct, but I don't see how it works. We can't just take $p-1 \to p$ as $p \to \infty$, can we?

The term $\frac{x_2^p}{x_1^{p-1}}=\left(\frac{x_2}{x_1}\right)^{p-1} x_{2}$. If $x_2$ is constant (which it is) and $-1<\frac{x_2}{x_1}<1$ then the term: $\left(\frac{x_2}{x_1}\right)^{p-1}$ still goes to zero.
I think the only issue you might have is when some of the $x_i$'s are negative (particularly if $|x_n| > |x_1|$). If you make the extra assumption that the values are non-negative then this looks ok.
• One could also change $p$ to $2p$, right? – MCT Jun 13 '16 at 17:50
• For negative? No, there is real trouble then in the denominator, $x_1=1$, $x_2=-1$. – André Nicolas Jun 13 '16 at 18:10
• @Soke, you could change the $p$'s to $2p$ and $p-1$ to $2p-1$, but I don't think that would solve the negativity problems. What you want to happen is if $x_1 > x_2$ then $\left(\frac{x_2}{x_1}\right)^p \to 0$. But this requires that the absolute value of $\frac{x_2}{x_1}$ to be smaller than 1. Simply making $x_1>x_2$ doesn't guarantee this. – TravisJ Jun 13 '16 at 18:48
Your RHS in the last equation should be like this (I think)$$\lim_{p \to \infty} \frac{\left({x_2\over x_1}\right)^{p-1}x_2+\dots+\left({x_n\over x_1}\right)^{p-1}x_n}{1+\frac{x_2^{p-1}}{x_1^{p-1}}+\dots+\frac{x_n^{p-1}}{x_1^{p-1}}}$$
Now each of the $\left({x_i\over x_1}\right)$ is less than $1$, So raised to power $\infty$ will make each of them zero and hence the numerator(each $x_i$ is finite ). And you are done I think..