Showing $2(\|f\|_{L_p}^p+\|g\|_{L_p}^p)\le\|f-g\|_{L_p}^p+\|f+g\|_{L_p}^p$

144 Views Asked by At

I have the following question from a past qualifying exam:

Given $2\le p<\infty$. Show that for any real-valued functions $f,g\in L_p(\mathbb{R})$, it holds that $$2\left(\left\|\frac{f}{2}\right\|_{L_p}^p+\left\|\frac{g}{2}\right\|_{L_p}^p\right)\le\left\|\frac{f-g}{2}\right\|_{L_p}^p+\left\|\frac{f+g}{2}\right\|_{L_p}^p\le\frac12(\|f\|_{L_p}^p+\|g\|_{L_p}^p).$$

The right inequality is given by Minkowski's inequality: $$\left\|\frac{f-g}{2}\right\|_{L_p}^p+\left\|\frac{f+g}{2}\right\|_{L_p}^p=\frac{1}{2^p}\left(\left\|f-g\right\|_{L_p}^p+\left\|f+g\right\|_{L_p}^p\right)\le\frac1{2^{p-1}}(\|f\|_{L_p}^p+\|g\|_{L_p}^p)\le\frac12(\|f\|_{L_p}^p+\|g\|_{L_p}^p).$$ For the left inequality, it suffices to show $$2(\|f\|_{L_p}^p+\|g\|_{L_p}^p)\le\|f-g\|_{L_p}^p+\|f+g\|_{L_p}^p.$$ How can this be done?

1

There are 1 best solutions below

0
On BEST ANSWER

Replacing the right-hand inequality: $(f,g)\to(\frac{f-g}{2},\frac{f+g}{2})$ gives $$ \left\lVert\frac{\frac{f-g}2+\frac{f+g}2}2\right\rVert_{L_p}^p + \left\lVert\frac{\frac{f-g}2-\frac{f+g}2}2\right\rVert_{L_p}^p \leq \frac12 \left( \left\lVert\frac{f-g}2\right\rVert_{L_p}^p + \left\lVert\frac{f+g}2\right\rVert_{L_p}^p\right) $$ or equivalently, $$ \left\lVert\frac{f}2\right\rVert_{L_p}^p + \left\lVert\frac{g}2\right\rVert_{L_p}^p \leq \frac12 \left( \left\lVert\frac{f-g}2\right\rVert_{L_p}^p + \left\lVert\frac{f+g}2\right\rVert_{L_p}^p\right) $$ which is the left-hand inequality when you move the 2 over.