prove the inequalities: $|x|_q \leq |x|_p \leq N^{^1/p-^1/q}\cdot |x|_q, x \in \mathbb{R}^{N}$ and $||x||_{l^q} \leq ||x||_{l^p}$ for $x\in l^p$

Question

I want to show two inequalities in normed spaces:

1.) $$|x|_q \leq |x|_p \leq N^{^1/p-^1/q} \cdot |x|_q, \hspace{3mm} x \in \mathbb{R}^{N}$$ and

2.)$$||x||_{l^q} \leq ||x||_{l^p},\hspace{5mm} x\in l^p$$

with $$1\leq p\leq q \leq \infty.$$

Edit: I screwed up the order of p and q in the inequality at first.

Answer 1

1. Let us prove that $\|x\|_q \leq \|x\|_p$ for every $x\in \ell^p$ (the proof is the same in $\mathbb{R}^n$).

Clearly it is enough to consider the case $x\neq 0$. As you have already observed, the proof is not difficult for a normalized vector. Hence, let us define $$ y := \frac{x}{\|x\|_p} $$ so that $\|y\|_p = 1$. In particular, $|y_k|\leq 1$ for every $k$, so that $|y_k|^q \leq |y_k|^p$ and $$ \|y\|_q = (\sum |y_k|^q)^{1/q} \leq (\sum |y_k|^p)^{1/q} = \|y\|_p^{p/q} = 1. $$ Hence, by the very definition of $y$, it follows that $\|x\|_q \leq \|x\|_p$.

2. Let us prove that $|x|_p \leq N^{1/p - 1/q} |x|_q$ for every $x\in \mathbb{R}^N$. Let us consider the exponent $\alpha := q/p > 1$ and its conjugate exponent $\alpha' = q/(q-p)$. By Holder's inequality we have that $$ \sum |x_k|^p \leq \left[ \sum (|x_k|^p)^\alpha \right]^{1/\alpha} N^{1/\alpha'} $$ i.e. $$ |x|_p^p \leq |x|_q^p N^{(q-p)/q} \quad \Longrightarrow \quad |x|_p \leq |x|_q N^{1/p - 1/q}. $$

Answer 2

1.) I recall the Power Mean inequality. Let $x_1,\dots, x_n$ be positive real numbers. The power mean of order $r$ is defined by

$M(x_1,\dots, x_n)(0) =\sqrt[n]{x_1\cdots x_n}$, $M(x_1,\dots,x_n)(r) =\left(\frac{x_1^r+\dots+x_n^r}{n}\right)^{\frac 1r}$ $(r\ne 0)$.

Then the function $M(x_1,\dots x_n):\Bbb R\to\Bbb R$ is continuous and non-decreasing. This claim should follow from the convexity of $x\ln x$ or the convexity of $x^\lambda$ $(\lambda\ge 1)$ and Jensen's inequality. When $r\ge 0$ then a similar claim holds for non-negative $x_1,\dots, x_n$.

Now let $x=(x_1,\dots, x_N)\in\Bbb R^N$. Put $x^*=(|x_1|,\dots, |x_N|)$.

Then $|x|_q=M(x^*)N^{-1/q}$, $|x|_p=M(x^*)N^{-1/p}$.

So $|x|_q N^{1/q}\le |x|_p N^{1/p}$.