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Here is a statement that I came upon whilst studying Sobolev spaces, which I cannot quite fill in the gaps:

If $s>t>u$ then we can estimate: \begin{equation} (1 + |\xi|)^{2t} \leq \varepsilon (1 + |\xi|)^{2s} + C(\varepsilon)(1 + |\xi|)^{2u} \end{equation} for any $\varepsilon > 0$ (here $\xi \in \mathbb{R}^n$ and $C(\varepsilon)$ is a constant, dependent on $\varepsilon$).

How can I show this? Many thanks for hints!

2 Answers 2

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We can write $t=\alpha u+(1-\alpha)s$, where 0<\alpha<1. We have to find $C(\varepsilon)$ that for each real number $x\geq 1$ we have $x^{2t}\leq \varepsilon x^{2s}+C(\varepsilon)x^{2u}.$ We apply Young's inequality to $ab\leq \frac{a^p}p+\frac{b^{p'}}{p'}$ for $a,b$ positive to $p=\frac 1{\alpha}$ (so $p'=\frac p{p-1}=\frac 1{1-\alpha}$). We get that \begin{align}x^{2t}&=x^{2\alpha u}\varepsilon^{1/p}p^{1/p} x^{2(1-\alpha)s}\varepsilon^{-1/p}p^{-1/p}\\\ &\leq \frac 1p x^{2\alpha up}\varepsilon p+(1-\alpha)x^{2s}\varepsilon^{-\alpha}p^{-p'/p}\\\ &=\varepsilon x^{2u}+C(\varepsilon)x^{2s}p^{-p'/p}, \end{align} with $C(\varepsilon)=(1-\alpha)\varepsilon^{-\alpha}$.

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Let $f(x)=(1+x)^{2(t-u)}-\varepsilon(1+x)^{2(s-u)}$. We have $f(0)=1-\varepsilon$ and since $s-u>t-u$ and $\varepsilon>0$, we know $f(x)\to-\infty$ as $x\to\infty$. Hence C(\varepsilon):=\displaystyle\sup_{0\leq x<\infty}f(x)<\infty.