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I am reading "Minimax Methods in Critical Point Theory with Applications to Differential Equations" from Rabinowitz. On page 31 it says

"By standard inequalities, since $\mu>2$, for $\varepsilon>0$, $$ \lVert u\rVert_{L^2(\Omega)}\leq a_8 \lVert u\rVert_{L^\mu(\Omega)}\leq a_9 K(\varepsilon)+\varepsilon \lVert u\rVert_{L^\mu(\Omega)}^\mu $$ where $K(\varepsilon)\to\infty$ as $\varepsilon\to0$."

The first inequality follows easily from Hölder inequality, however I have never heard of the second. How do you proof the second inequality?

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My guess is this is Young's inequality, $$ ab \leq \frac{a^p}{p}+\frac{b^q}{q} $$ ($1/p+1/q=1$, $a,b\geq 0$). If we choose $b=(\mu \varepsilon)^{1/\mu}\lVert u \rVert_{\mu}$, $q=\mu$, $a=a_8(\mu \varepsilon)^{-1/\mu}$, this turns into $$ a_8 \lVert u \rVert_{\mu} \leq \frac{\mu-1}{\mu} \left(a_8(\mu \varepsilon)^{-1/\mu}\right)^{\mu/(\mu-1)} + \varepsilon \lVert u \rVert_{\mu}^{\mu}; $$ in particular, the power of $\varepsilon$ is $-1/(\mu-1)<0$.

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Chappers beat me to an answer, but since my approach is slightly different I figured I'd post it also.

It suffices to show that for all $\varepsilon > 0,$ there is $K(\epsilon) > 0$ such that for all $a>0,$ we have $a \leq K(\varepsilon) + \varepsilon a^{\mu}.$ To show this, note that $\varepsilon a^{\mu} - a \rightarrow \infty$ as $a \rightarrow \infty$ (since $\mu > 2$), so there is $K(\epsilon)>0$ such that $\epsilon a^{\mu} > a$ whenever $a > K(\epsilon).$ Note this gives the desired $K(\epsilon).$

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  • $\begingroup$ @Chappers Whoops, I actually meant $\mu > 2$ since that was the assumption we had. Thanks for the correction. $\endgroup$ – ktoi Jun 12 '17 at 17:17

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