# Show that some function is in $L^p$ space

lebesgue-measurelp-spacesmeasure-theoryreal-analysis

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 $$x^a and we can get $$p \in (\frac{1}{a},\infty)$$.

Then, since $$a, we know that $$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?

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 $$x \to \infty$$.