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Prove that if $x,y \in R$, and $1 \leq p < \infty$ then $|x+y|^p \leq 2^p(|x|^p+|y|^p )$

what I'm thinking is

$|x+y|\leq |x|^p+|y|\le 2\cdot\max\{|x|,|y|\}$

Without loss of generality , let $\max\{|x|, |y|\}=|x|$

hence and $|x+y|\leq 2|x| \Rightarrow |x+y|^p \le (2|x|)^p = 2^p|x|^p $

$|x+y|^p \le 2^p|x|^p $ and since $|y| \ge 0 $, we have

$ |x+y|^p \le 2^p(|x|^p+|y|^p )$

Is there any idea I could have used here?

Martin Sleziak
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Cnine
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2 Answers2

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In fact, it is not hard to see that $\left(\max\{|x|,|y|\}\right)^{p}=\max\{|x|^{p},|y|^{p}\}\leq|x|^{p}+|y|^{p}$. Combining together with $|x+y|^{p}\leq(|x|+|y|)^{p}\leq\left[2\max\{|x|,|y|\}\right]^{p}$ yields the result.

user284331
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Hint: use the convexity of the function $h(x)=x^p$ for $x\geq0$ and $p>1$. Apply the convexity on $\Big|\frac{1}{2}|f|+\frac{1}{2}|g|\Big|^p$

Isko10986
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  • okay, the extra information is helpful – Cnine Dec 01 '16 at 03:20
  • So $\left|(1/2)|f|+(1/2)|g|\right|^{p}\leq(1/2)|f|^{p}+(1/2)|g|^{p}$, multiplying both sides by $2^{p}$ we get $(|f|+|g|)^{p}\leq 2^{p-1}(|f|^{p}+|g|^{p})$, we obtain a better constant. – user284331 Dec 01 '16 at 03:59
  • If I am right, $h(x)$ is convex if for $0 \le \alpha \le 1, \quad h(\alpha x +(1-\alpha )y \le \alpha h(x)+(1-\alpha )h(y)$, given $h(x)=|x|^p$

    So, if we let $\alpha =\frac{1}{2}. \quad h(\frac{1}{2}|x|+\frac{1}{2}|y|)\le \frac{1}{2}h(|x|)+\frac{1}{2}h(|y|)$ $\Rightarrow |\frac{1}{2}|x|+\frac{1}{2}|y||^p \le \frac{1}{2}|x|^p+\frac{1}{2}|y|^p$

    therefore $(\frac{1}{2})^p | |x|+|y||^p \le \frac{1}{2}(|x|^p+|y|^p)$ so $| |x|+|y||^p \le 2^{p-1}(|x|^p+|y|^p)$

     – Cnine Dec 01 '16 at 04:02
  • oh okay, I know is true regardless, since $p\ge 1$. but is there a way to get $2^p$ instead of $2^{p-1}$? – Cnine Dec 01 '16 at 04:05
  • multiply the right side by 2. – Isko10986 Dec 01 '16 at 04:14