For $0 < a < b < \infty$, define $f:(0,\infty) \rightarrow \mathbb{R}$ as
$$ f(x) = \frac {1} { x^a + x^b } ; x>0$$
Find all $p \geq 1 $ such that $f \in L^p ((0,\infty))$.
My solution:
First, I separate interval in $(0,1) , [1,\infty)$
(i) when $x \in (0,1)$
Then, $|f(x)| = \frac {1} {x^a+x^b} \leq \frac{1}{2x^b}$ and by using this inequality, we can get
$\int_0^1 |f(x)|^p dx \leq \int_0^1 (\frac{1}{2x^b})^p dx = \frac{1}{2^p} \int_0^1 x^{-bp} dx = \frac{1}{2^p} \cdot \frac{1}{-bp+1} \cdot x^{-bp+1} |^1_0$.
To bound the above integral, $p$ has to be in $(\frac{1}{b},\infty)$.
(ii) $x \in [1,\infty)$
In same way, use that $x^a <x^b$ and we can get $p \in (\frac{1}{a},\infty)$.
Then, since $a<b$, we know that $f \in L^p((0,\infty))$, when $p \in (\frac{1}{a},\infty)$.
Does my solution need any correction? I think I might miss some details.
If so, how could I solve this problem?
Best Answer
Your first computation is right, but the conclusion is wrong. It should be $bp < 1$ so that $p < \frac{1}{b}$. Your second conclusion $p > \frac{1}{a}$ is correct. So from your computation, you can conclude that the valid range of $p$ contains $(\frac{1}{a}, \frac{1}{b}) = \emptyset$. But your argument does not allow you to conclude that the valid range of $p$ actually equals $(\frac{1}{a}, \frac{1}{b})$. From my computations, the answer is $(\frac{1}{b}, \frac{1}{a})$. The key is to observe that $\frac{1}{x^a + x^b} \sim \frac{1}{x^a}$ as $x \to 0$ and that $\frac{1}{x^a + x^b} \sim \frac{1}{x^b}$ as $x \to \infty$.