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I am reading the book "Elliptic Partial Differential Equations of Second Order" by D. Gilbarg and N.S. Trudinger.

In theorem 7.27, it stated that after we obtained $$|u'(x)| \leq \int_{a}^{b}|u''| + \frac{18}{\epsilon^2}\int|u|,$$ we can apply Holder's inequality to get $$|u'(x)|^p \leq 2^{p-1}\left\{ \epsilon^{p-1}\int_{a}^{b}|u''|^p + \frac{(18)^p}{\epsilon^{p+1}}\int_{a}^{b}|u|^p \right\}.$$

I am wondering how Holder's inequality is used to get the second inequality.

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In that proof, $\epsilon=b-a$. So,

$$\begin{align} \int_{a}^{b}|u''|&=\int_{a}^{b}|1\cdot u''|\\ &\leq\left(\int_{a}^{b}|1|^q\right)^{1/q}\left(\int_{a}^{b}|u''|^p\right)^{1/p}\\ &=(b-a)^{1/q}\left(\int_{a}^{b}|u''|^p\right)^{1/p}\\ &=\epsilon^{(p-1)/p}\left(\int_{a}^{b}|u''|^p\right)^{1/p} \end{align}$$

Now, do a similar calculation for the integral

$$\frac{18}{\epsilon^2}\int|u|=\frac{1}{\epsilon^2}\int|1\cdot 18u|,$$

take the power $p$ and use the inequality $$(a+b)^p\leq 2^{p-1}(a^p+b^p).$$

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  • $\begingroup$ I get it now. Thank you very much! $\endgroup$
    – Sum
    Oct 24, 2015 at 1:56

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