I have no idea how to show this in the case that $0<p<1$. I can show that $|\alpha-\beta|^p\leq 2^{p-1}(|\alpha|^p+|\beta|^p)$ for $p\geq1$ by using the convexity of the function $x^p$, but am lost at how to solve for this specific case. Any guidance would be much appreciated. Thanks!


$$|\alpha - \beta|^p \leq (|\alpha| + |\beta|)^p$$

We want to show $$(|\alpha| + |\beta|)^p \leq |\alpha|^p + |\beta|^p$$

which is equivalent to $$1 \leq (\dfrac{|\alpha|}{|\alpha| + |\beta|})^p + (\dfrac{|\beta|}{|\alpha| + |\beta|})^p$$

when $0< \lambda<1$ and $0 < p<1$, we have $\lambda^p > \lambda$, so $$(\dfrac{|\alpha|}{|\alpha| + |\beta|})^p \geq \dfrac{|\alpha|}{|\alpha| + |\beta|}$$ and $$(\dfrac{|\beta|}{|\alpha| + |\beta|})^p \geq \dfrac{|\beta|}{|\alpha| + |\beta|}$$

  • $\begingroup$ Thank you! I always hate how simple problems are in hindsight... $\endgroup$ – PiTheMathemagician Oct 1 '14 at 15:58

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