If $\int_{1}^{\infty} |f(t)|^{2}\,dt < \infty$, is $\int_{1}^{\infty} \frac{|f(t)|}{\sqrt{t}}\,dt < \infty$?

Question

If $\int_{1}^{\infty} |f(t)|^{2}\,dt < \infty$, is $\int_{1}^{\infty} \frac{|f(t)|}{\sqrt{t}}\,dt < \infty$?

If Yes, give a proof. If no, give a counterexample.

I try to use Holder inequality to prove, but I couldn't prove and also I couldn't find counterexample. Anyone can help me?

Answer 1

The statement is false. Take $$f(t)=\frac{1}{\sqrt{t}\log(1+ t)}.$$ Then $$\int_1^\infty|f(t)|^2\,dt=\int_1^\infty\frac{1}{t\log^2(1+t)}<\infty$$ while we have on the other hand $$\int_1^\infty \frac{|f(t)|}{\sqrt{t}}\,dt=\int_1^\infty\frac{1}{t\log(1+t)}\,dt=\infty.$$