Is it $L^2$ function ?

Question

For $f ∈ L^2 ([0, 1))$, define $Vf(x) = \int_{0}^{x}f(t) dt$. Is $Vf$ continuous function? Is it in $L^2$?

Thanks for any help

Answer 1

The answer is yes to both questions: Let $x,y\in [0,1)$ such that $y<x$. Then $$ |Vf(x)-Vf(y)|=|\int_y^xf(t)dt|\leq\int_y^x|f(t)|dt=\int_0^1|f(t)|\chi_{[y,x]}(t)dt\leq ||f||_2|x-y|^{1/2} $$ where in the last inequality we use the Cauchy-Schwartz inequality. So $Vf$ is continuous. On the other hand, by the previous step $$ \int_0^1|\int_0^xf(t)dt|^2dx\leq\int_0^1||f||_2^2xdx=\frac{||f||_2^2}{2} $$ So $Vf$ is in $L^2[0,1)$.