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I'm reading Folland's Introduction to Partial Differential Equations, and he makes a few claims that I don't understand and I think may be typos in the book.

Firstly let's fix $$ L = \sum_{|\alpha|=k}a_\alpha \partial^{\alpha} $$ and assumme the $a_\alpha$ are constants (Folland does it in more generality, but I just want to look at the constant case)

In the following proof, he computes $$\widehat{Lu}(\xi) = (2\pi i)^k \sum_{|\alpha|=k}a_\alpha\xi^\alpha \widehat{u}(\xi)$$ and then claims that by the assumed ellipticity condition $$ \left|\sum_{|\alpha|=k}a_\alpha \xi^\alpha \right| \geq A|\xi|^k $$ we have $$ (1+|\xi|^2)^s|\widehat{u}(\xi)|^2 \leq 2^k(1+|\xi|^2)^{s-k}(1+|\xi|^2)^k|\widehat{u}(\xi)|^2 $$ $$ \leq 2^k A^{-1}(1+|\xi|^2)^{s-k}|\widehat{Lu}(\xi)|^2+2^k(1+|\xi|^2)^{s-k}|\widehat{u}(\xi)|^2. $$

I have been fiddling with it for a few hours now and I have no idea how the second inequality follows. I suspect the $2^k$ in the first was a typo (it would be equality without it), so maybe he was trying to claim $$ (1+|\xi|^2)^{s-k}(1+|\xi|^2)^k|\widehat{u}(\xi)|^2 \leq 2^k A^{-1}(1+|\xi|^2)^{s-k}|\widehat{Lu}(\xi)|^2+2^k(1+|\xi|^2)^{s-k}|\widehat{u}(\xi)|^2 $$ but I have no idea what he did in order to reach the conclusion. Could someone enlighten me as to what Folland meant to write?

nullUser
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1 Answers1

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I figured it out, he was using $(1+|x|)^k \leq 2^k(1+|x|^k)$. It is easy to check for $|x|<1, |x|=1$ and $|x|>1$.

nullUser
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  • It should be noted that you can improve this inequality a little to $(1 + | x |)^k \le 2^{k - 1}(1 + | x |^k)$ for $k \in [0, \infty)$ by noting the convexity of $t \mapsto t^{k}$ which, by Jensen's inequality, gives $$\left( \frac{1 + | x |}{2}\right)^k \le \frac{1 \cdot 1^k + 1 \cdot | x |^k}{2} = \frac{1 + | x |^k}{2},$$ which now only has to be rearranged. – ViktorStein Nov 22 '19 at 19:09