I discovered last night that I have an error in my proof to the following problem and I need help fixing it (or need a new solution) $$ \text{Suppose that} \int_{0}^\infty x^{-2}|f|^5 dx < \infty. \text{Prove that} \lim_{t \to 0} t^{-\frac{6}{5}} \int_0^t f(x)dx = 0 $$
Here is what I tried:
Since the integrand of the first integral is always positive, $\int_0^t x^{-2}|f|^5 < \infty$, so $x^{-2/5}|f| \in L^5$. Since $\mu{[0,t]}$ is finite, $x^{-2/5}|f| \in L^4$ as well.
Thus,
$t^{-6/5}\int_{[0,t]} f(x)dx = t^{-6/5}\int \frac{x^{2/5}}{x^{2/5}}f(x)\chi_{[0,t]}^2 dx \leq t^{-6/5}||x^{-2/5}f\chi_{[0,t]}^2||_4||x^{2/5}\chi_{[0,t]}||_{4/3}$
At this point I make a computation error and so my proof breaks down. The rough idea is that the last norm can be evaluated as a definite integral to get $t$ raised to some power. The positive powers of $t$ should cancel with the $t^{-6/5}$ so that we get a constant times $t$ raised to a positive power. Then we can let $t$ go to $0$ to get the result.
If you're not going to $L^4$ (since there is no good reason to...), in the last inequality you get $$ t^{-6/5}\|x^{-2/5}f\chi_{[0,t]}\|_5\|x^{2/5}\chi_{[0,t]}\|_{5/4}. $$ A simple integration gives $\|x^{2/5}\chi_{[0,t]}\|_{5/4}=t^{6/5}(2/3)^{4/5}$, so you get $$ (2/3)^{4/5}\|x^{-2/5}f\chi_{[0,t]}\|_5, $$ which goes to $0$ indeed, by DCT.