Let $a, b$, be positives and $a \leq b$. For which $p$ the function $\frac{1} {x^a+x^b}$ is $L_p(0,+\infty)$?

Question

Let $a, b$, be positives and $a \leq b$. For which $p$ the function $\frac{1} {x^a+x^b}$ is $L_p(0,+\infty)$?

I have solved without any problem the case $p=\infty$ (and it is not $L_p$ in this case) , but I can't solve the other case.

I suppose I have to use some inequalities but I have no idea which ones, and how.

Anyone can help me?

Answer 1

Since $b\geq a>0$ you will have $$ \frac{1}{x^a+x^b}\approx \frac{1}{x^a}\quad\text{if }x\approx 0 $$ and $$ \frac{1}{x^a+x^b}\approx \frac{1}{x^b}\quad\text{as }x\to+\infty. $$ Now use comparison to make this argument complete.

Answer 2

For $x \in (0,1)$ we have $$x^a \le x^a+x^b \le 2x^a$$ so $$\int_{(0,1)}\frac1{2^px^{ap}} \le \int_{(0,1)}\frac1{(x^a+x^b)^p} \le \int_{(0,1)}\frac1{x^{ap}}$$ Hence $\frac1{x^a+x^b}$ is in $L^p(0,1)$ if and only if $ap > 1$, or $p > \frac1a$.

Similarly, for $x \in (1, \infty)$ we have $$x^b \le x^a+x^b \le 2x^b$$ so $$\int_{(1, \infty)}\frac1{2^px^{bp}} \le \int_{(1, \infty)}\frac1{(x^a+x^b)^p} \le \int_{(1, \infty)}\frac1{x^{bp}}$$ Hence $\frac1{x^a+x^b}$ is in $L^p(1, \infty)$ if and only if $bp < 1$, or $p < \frac1b$.

Conclusion: $\frac1{x^a+x^b}$ is in $L^p(0, \infty)$ if and only if $$p \in \left(\frac1a, \frac1b\right)$$