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While it's a well known that any two norms are equivalent for a finite dimensional normed linear space, I've been trying to derive the bounds for the case $X=\mathbb{R}^n$ and $l_p$-norms.

Let $1 \leq p,q <\infty.$ If $q>p$, from Holder's inequality we can derive that $$ \|x\|_p \leq (n)^{1/p-1/q}. \|x\|_q $$

But I am unable to derive a similar $\leq$ tightest bound for the case $q<p$. Can anyone please provide a hint for this ?

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We have $ \lVert x \rVert_p \leq \lVert x \rVert_1$ for all $1 \leq p < \infty $.

Why? Because $|x_1|^p + \ldots + |x_n|^p \leq (|x_1| + \ldots + |x_n|)^p$ (from Newton's binomial formula). So $(|x_1|^p + \ldots + |x_n|^p)^{1/p} \leq |x_1| + \ldots + |x_n|$.

For given $1 \leq q < \infty $ we have: $|1 \cdot x_1| + \ldots + |x_n \cdot 1| \leq |1+ \ldots + 1|^{(1-1/q)} (|x_1|^q + \ldots + |x_n|^q)^{1/q} $

from Holder's formula.

Hence

$(|x_1|^p + \ldots + |x_n|^p)^{1/p} \leq |x_1| + \ldots + |x_n| \leq n^{(1-1/q)} (|x_1|^q + \ldots + |x_n|^q)^{1/q} \implies \lVert x \rVert_p \leq n^{(1-1/q)} \lVert x \rVert_q$

which is true for all $1 \leq p,q < \infty $.

If $p > q$, the tighest bound is $1$: $ \lVert x \rVert_p \leq \lVert x \rVert_q$

Why? we need:$ \lVert x \rVert_p \leq \lVert x \rVert_q$ by dividng both sides by $\lVert x \rVert_q$ we derive:

$\lVert \frac{x}{\lVert x \rVert_q} \rVert_p \leq 1 $

$ (| \frac{x_1}{\lVert x \rVert_q}|^p + \ldots + |\frac{x_n}{\lVert x \rVert_q}|^p)^{1/p} \leq 1 $

$ | \frac{x_1}{\lVert x \rVert_q}|^p + \ldots + |\frac{x_n}{\lVert x \rVert_q}|^p \leq 1 $

$ | \frac{x_1}{\lVert x \rVert_q}|^p + \ldots + |\frac{x_n}{\lVert x \rVert_q}|^p \leq | \frac{x_1}{\lVert x \rVert_q}|^q + \ldots + |\frac{x_n}{\lVert x \rVert_q}|^q = 1$.

This boundary couldn't less than $1$ beacuse for $x=(1,0,\ldots,0)$ we have equality: $ \lVert x \rVert_p = \lVert x \rVert_q$.

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  • $\begingroup$ Is the above bound tight ? I want a tightest bound. $\endgroup$ Commented Jun 23, 2015 at 5:57
  • $\begingroup$ Oh, I thought you're looking for both $p \leq p$ and $q \leq p$ cases, sorry. $\endgroup$
    – ninjaaa
    Commented Jun 23, 2015 at 6:25
  • $\begingroup$ No problem. Anyways, thanks for the above insight. $\endgroup$ Commented Jun 23, 2015 at 7:01
  • $\begingroup$ I found the minimal boundary in this case and make update. $\endgroup$
    – ninjaaa
    Commented Jun 23, 2015 at 8:18
  • $\begingroup$ Thanks a lot. I failed to notice this simple observation. $\endgroup$ Commented Jun 23, 2015 at 10:25

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