I am trying to understand the properties of Lebesgue integration, and in particular how exponents play between the integrand and the value of the integral.
For instance, suppose that it is known $$\int_{[0,1]} f_n(x)^2 d\lambda \leq \frac{1}{n^4}$$ for all $n \in \mathbb{N}$, where $\lambda$ is the Lebesgue measure on $[0,1]$.
It seems to me, since $+\sqrt{f_n(x)^2} = |f_n(x)|$ for every $x$ (where we consider only the positive values of $\sqrt{f_n(x)^2}$) then we should have that $$\int_{[0,1]} |f_n(x)| d \lambda \leq \frac{1}{n^2}.$$ I am attempting to prove this, but I am not sure how.
I realize that in the $L^2$ norm, the statement $||f_n||_2 \leq \frac{1}{n^2}$ holds (since $f_n^2 = |f_n|^2$), but it is not in general true that $$\int|f_n(x)|^2 d \lambda = \left( \int |f_n(x)| d \lambda \right )^2.$$ Would anyone be able to provide a hint on how to bound the statement above, or hint at a counterexample?
Thank you!
Use Schwarz's inequality: $$ \int_{[0,1]} |f_n(x)|\, d\lambda\le\Bigl(\int_{[0,1]} |f_n(x)|^2\, d\lambda\Bigr)^{1/2}\Bigl(\int_{[0,1]} d\lambda\Bigr)^{1/2}. $$