Find $f\in L^p(0,1)$, but $f\notin L^q(0,1)$, $p < q$

Question

I want to find a measurable function $f \in L^p(0,1)$, for $p\in [1,+\infty)$, but $f \notin L^q(0,1)$ for each $q\in (p,+\infty]$.

I tried to manipulate $f=\frac{1}{x^a}$, improving the exponent $a$ s.t. $\int_0^1 |\frac{1}{x^a}|^p d\mu < \infty $, but since $p,q$ are real number, I cannot find anything - I don't know how to say it properly, but I cannot find an explicit expression s.t. some algebraic tricks make appear something smaller or bigger than 1, which is the limit point for the divergence and the convegercence of the function- (otherwise, if we deal with the integers I found $ f=\frac{1}{x^{\frac{q-1}{p^2}}} $, where $q \geq p$ )

Can anyone help me? Thanks in advance.

Answer 1

You can try with the function $$ f(x) = \begin{cases} (x \log^2 x)^{-1/p}, &\text{if}\ x\in (0,1/2),\\ 0, &\text{if}\ x \in (1/2, 1). \end{cases} $$