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I study Trudinger's inequality and the following inequality appears:

Let N be a natural number. For each $\epsilon>0, \exists C_\epsilon$ such that $1+s^{(N-1)/N} \leq ((1+\epsilon)s+C_\epsilon)^{(N-1)/N} \forall s\geq 0$

I tried to differentiate the function $(RHS)-(LHS)$ to deduce this is increasing, but it does not work because out exponent is smaller than $1$. So How can I get it?

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1 Answer 1

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Let $p = \frac{N-1}{N}$, and consider the function

$$g_\epsilon \colon s \mapsto \bigl(1 + s^p\bigr)^{1/p} - (1+\epsilon)s.$$

Then differentiating yields

\begin{align} g_\epsilon'(s) &= \frac{1}{p}\bigl(1 + s^p\bigr)^{\frac{1}{p}-1}\cdot ps^{p-1} - (1+\epsilon) \\ &= \bigl(1 + s^p\bigr)^{\frac{1}{N-1}}\cdot s^{-\frac{p}{N-1}} - (1 + \epsilon) \\ &= \bigl(1 + s^{-p}\bigr)^{\frac{1}{N-1}} - (1+\epsilon) \end{align}

for $s > 0$. Since $g'_\epsilon(s) > 0$ for small enough $s$, and $\lim\limits_{s\to +\infty} g'_\epsilon(s) = -\epsilon < 0$, it follows that $g_\epsilon$ attains its maximum at some $s_0 > 0$. Then choosing any

$$C_\epsilon \geqslant g_\epsilon(s_0)$$

works.

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  • $\begingroup$ Oh, so it is just a calculus TT... Thank you a lot! $\endgroup$
    – CSH
    Nov 21, 2016 at 14:04

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