$||x||_p$ norms are equivalent on $\mathbb{R}^n$ for $1 \leq p \leq \infty$. I have a proof that any of the $||x||_p$ norms on $\mathbb{R}^n$ are equivalent, and I just want to be sure I haven't made a mistake anywhere and if there are some improvements. Thanks in advance!
So first note that if $M = \max_{1 \leq i \leq n} |x_i| = ||x||_\infty$
\begin{align*}
 |x_1|^p+...+|x_n|^p &\leq M^p+...+M^p = nM^p \\
||x||_p = \left(|x_1|^p+...+|x_n|^p\right)^{\frac{1}{p}} &\leq n^{\frac{1}{p}}M = n^{\frac{1}{p}} ||x||_\infty
 \end{align*}
Next, assume without loss of generality that $|x_j|, \; 1\leq j \leq n$ corresponds to the infinity norm of $x$. Then
\begin{align*}
  |x_j|^p &\leq |x_1|^p+...+|x_j|^p+...+|x_n|^p \\
 ||x||_\infty = |x_j| &\leq \left( |x_1|^p+...+|x_j|^p+...+|x_n|^p \right)^{\,\frac{1}{p}} = ||x||_p
 \end{align*}
Therefore if we combine these two inequalities we get
\begin{align}
||x||_\infty \leq ||x||_p \leq n^{\frac{1}{p}} ||x||_\infty
\end{align}
Which implies
$$
\frac{1}{n^{\frac{1}{p}}}\leq  \frac{||x||_p}{||x||_\infty} \leq n^{\frac{1}{p}}
$$
So for any $1\leq p <q$ we can use the above inequality twice to get
$$
\frac{1}{n^{\frac{1}{p}}} \frac{1}{n^\frac{1}{q}} \leq  \frac{||x||_p}{||x||_\infty} \frac{||x||_\infty}{||x||_q} \leq n^{\frac{1}{p}} n^{\frac{1}{q}}
$$
And so
$$
\frac{1}{n^{\frac{1}{p}+\frac{1}{q}}} \leq  \frac{||x||_p}{||x||_q} \leq n^{\frac{1}{p}+\frac{1}{q}}
$$
And the norms are equivalent.
 A: Someone should give a proof that any two norms on $\Bbb R^n$ are equivalent.
As suggested, we assume that $||\cdot||$ is a norm and try to show it's equivalent to $||\cdot||_\infty$. There's a getting-started problem here: Dealing with the $\ell_p$ norms we just play with the coordinates, but given only that $||\cdot||$ is a norm it's not at all clear at first how $||x||$ has anything to do with $|x_j|$.
Getting Started: Say $e_1,\dots,e_n$ is the standard basis for $\Bbb R^n$. Note that saying $x=(x_1,\dots,x_n)$ is the same as saying $$x=\sum_jx_je_j.$$
That's almost half the proof, which is to say if you were going to do it yourself this would be a reasonable place to stop reading and resume scribbling...
Now the triangle inequality shows that $$||x||\le\sum|x_j|\,||e_j||\le\left(\sum||e_j||\right)||x||_\infty=c||x|_\infty,$$and we're exactly half done.
Again, if you want to do it yourself, the other inequality follows by a simple compactness argument.
Since $||x||\le c||x||_\infty$ it follows that the map $x\mapsto||x||$ is continuous on $\Bbb R^n$ (with the standard topology). Let $$S=\{x:||x||_\infty=1\}.$$Then $S$ is compact, and so by continuity there exists $\delta>0$ with $$||x||\ge\delta\quad(x\in S).$$That is, $$||x||\ge\delta\quad(||x||_\infty=1),$$which implies by homogeneity that $$||x||\ge\delta||x||_\infty.$$
