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Suppose $x\geq0$, $\epsilon>0$ and $p\in (1,\infty)$. How can i find the best constant ( if it exists!) $C>0$ in the inequality ($p$ and $\epsilon$ fixed):$$\frac{(\epsilon^2+x^2)^\frac{p-2}{2}}{\epsilon^{p-2}+x^{p-2}}\leq C,\ x\geq 0$$

Thanks

Edit: i haev made some changes.

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See the book of differential and integral calculus written by George F. Simmons Calculus with Analytical Geometry In the chapter on Lagrange multipliers is a section that tells how to apply Lagrange multipliers for inequalities like this property that you want. –  Elias Nov 14 '12 at 11:11

3 Answers 3

up vote 3 down vote accepted

The inequality is equivalent to $(1+t)^q\leqslant C\cdot(1+t^q)$, where $q=\frac12(p-2)$ hence $q\gt-\frac12$, and $t=x^2/\epsilon^2$ hence $t\gt0$. This cannot hold simultaneously for every $(t,p)$ since, for every fixed $t\gt0$, $(1+t)^q/(1+t^q)\to+\infty$ when $q\to+\infty$.

For every fixed $p$, let $C(p)$ denote the optimal constant such that this holds for every positive $t$. Then $$C(p)=\left\{\begin{array}{cl}1 & \text{if}\ 2\lt p\leqslant4,\\ 2^{(p-4)/2}& \text{otherwise.} \end{array}\right. $$ Note the discontinuity at $p=2$, from $C(2)=C(2^-)=\frac12$ to $C(2^+)=1$.

To prove this, consider the function $u_q:t\mapsto q\log(1+t)-\log(1+t^q)$ defined on $t\geqslant0$, then $u'_q(t)$ has the sign of $q\cdot(1-t^{q-1})$ hence $u_q(t)$ is maximal at $t=1$ if $q\gt1$ or $q\lt0$ and $u_q(t)$ is maximal at $t=0$ or $t=+\infty$ if $0\lt q\lt1$. The maximum is $(q-1)\log2=\frac12(p-4)\log2$ in the first case and $0$ in the second case. The values of $C(p)$ follow.

Edit: The OP asks in a comment whether, for some $p\gt1$ and $\epsilon\gt0$ fixed, there exists a finite constant $C_{p,\epsilon}$ such that $(\epsilon^2+x^2)^{(p-2)/2}\leqslant C_{\epsilon,p}+x^{p-2}$ for every positive $x$. The same approach as above yields the answer.

By homogeneity, $C$ is appropriate if and only if $C=D\,\epsilon^{p-2}$ where $(1+t)^q\leqslant D+t^q$ for every $t\geqslant0$, once again with $q=\frac12(p-2)$. Considering the function $v_q:t\mapsto(1+t)^q-t^q$ and its derivative, one sees that the supremum is $v_q(+\infty)=+\infty$ when $q\gt1$, $v_q(0)=1$ when $0\lt q\leqslant1$, $v_0(t)=0$ for any $t\gt0$ when $q=0$, and $v_q(+\infty)=1$ when $q\lt0$.

Thus, $C_{p,\epsilon}$ does not exist when $p\gt4$, $C_{p,\epsilon}=\epsilon^{p-2}$ when $p\leqslant4$ and $p\ne2$, and $C_{2,\epsilon}=0$. Finally, the optimal inequality one can achieve is the well-known fact that, for every $p\leqslant4$, $$ (\epsilon^2+x^2)^{(p-2)/2}\leqslant \epsilon^{p-2}+x^{p-2}. $$

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$p$ and $\epsilon$ are fixed, sorry about it. –  Tomás Nov 14 '12 at 11:34
    
I know. $ $ $ $ –  Did Nov 14 '12 at 11:35
    
@did Please give some details. How could you derive the $C_p$? (An earlier version was different.) –  vesszabo Nov 14 '12 at 11:39
    
@did, can i improve the term $\epsilon^{p-2}$ in such a way that $C_p=1$ for all $p$? –  Tomás Nov 14 '12 at 11:44
    
@vesszabo Done. –  Did Nov 14 '12 at 14:39

I don't think such a constant exists, if we allow $x$, $\epsilon$ and $p$ to all vary.

First, just a general trick: if you're dealing with multiple real numbers you can express them in terms of themselves and often reduce the number of variables. In this case fix $\epsilon$ and let $x=a\epsilon$, for $a=x/\epsilon \in[0,\infty)$. Then your quantity becomes $$ Q(\epsilon,x,p) := \frac{ ( \epsilon^2 + x^2 )^{(p-2)/2} }{ \epsilon^{p-2}+x^{p-2} } = \frac{ ( \epsilon^2 + (a\epsilon)^2 )^{(p-2)/2} }{ \epsilon^{p-2}+(a\epsilon)^{p-2} } = \frac{ ( 1 + a^2 )^{(p-2)/2} }{ 1+a^{p-2} } $$ because all the $\epsilon$ terms cancel out. Now if I write $q=(p-2)/2 \in(-1/2,\infty)$ just to clean things up I get $$ Q(a,q) = \frac{(1+a^2)^q}{1+a^{2q} }. $$

Now this is where I think the constant doesn't exist. Set $a=2$ to get $$ Q(2,q) = \frac{ (5)^q }{ 2^{2q} + 1 } = \frac{5^q}{4^q+1} \geq \frac{5^q}{4^q+4^q} $$ when $q>1$. Then $$ Q(2,q) = \frac{ 5^q}{ 2 \times 4^q} = \frac{1}{2} \left( \frac{5}{4} \right)^q $$ and so $$ \lim_{q\rightarrow \infty} Q(2,q) = + \infty $$ Thus no inequality of the form $Q(x,\epsilon,p) \leq C$ can exist.

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In short, let $x=1$, $\epsilon=2$ and take the limit as $p\rightarrow \infty$: wolframalpha.com/input/… –  James Fennell Nov 14 '12 at 11:29
1  
Sorry James, I think i cause some misunderstanding here. $p$ and $\epsilon$ is fixed and $x$ vary. –  Tomás Nov 14 '12 at 11:30
    
@Tomás - no problem! –  James Fennell Nov 14 '12 at 19:22

Using the inequality $$ (\epsilon^2+x^2)^{1/2}\le\epsilon+x, $$ the problem reduces to finding $C$ such that $$ (\epsilon+x)^{p-2}\le C\bigl(\epsilon^{p-2}+x^{p-2}\bigr). $$ Let $t=x/\epsilon$. Dividing by $\epsilon^{p-2}$ the last inequality is equivalent to $$ (1+t)^{p-2}\le C\bigl(1+t^{p-2}\bigr),\quad t\ge0. $$ From here it is easy to show that $C$ exists. For instance, if $p>2$, then the function $(1+t)^{p-2}/(1+t^{p-2})$ is continuous on $[0,\infty)$ and converges to $1$ as $t\to\infty$, so that it is bounded. You can use calculus to find $C$. Similarly if $1<p<2$.

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One loses too much in the first inequality to expect reaching the optimal constants afterwards. –  Did Nov 14 '12 at 14:41

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