Is 'f' belong sobolev?

Question

I was trying to show that the function

$$f(x) = \dfrac{x^{1/2}}{1+x^2} \in W^{1,3/2} (0,\infty)$$

that is, have to show that $$f\in L^{3/2}(0,\infty)$$ and $$f_x\in L^{3/2}(0,\infty).$$

Answer 1

Note that $$f_x = \frac{(1 + x^2) \frac 1 2 x^{-1/2} - x^{1/2} \cdot 2x}{(1 + x^2)^2} = \frac{\frac 1 2 x^{-1/2} - \frac 3 2 x^{3/2}}{(1 + x^2)^2}$$

For large values of $x$, we have the estimate (where $\lesssim$ means "up to some constants")

$$|f_x(x)| \lesssim \frac{x^{3/2}}{x^4} = x^{-5/2}$$

For small values of $x$, we have the similar estimate

$$|f_x(x)| \lesssim \frac{x^{-1/2}}{1} \implies |f_x(x)|^{3/2} \lesssim x^{-3/4}$$

Can you justify these, and use them to decide whether $f \in W^{1, 3/2}(0, \infty)$?