Norm of operator $\phi(f)=\int_{0}^{1}f(x^2)dx$ Does the map $\phi :L^3([0,1],\mathbb{C})\to \mathbb{C}$ such that $$\phi(f)=\int_{0}^{1}f(x^2)dx$$ define a bounded linear functional on the Banach space $L^3([0,1],\mathbb{C})?$ If yes, determine its norm.
Here linearity is obvious. i.e. $\phi(af+bg)=\int_{0}^{1}(af+bg)(x^2)dx=a\int_{0}^{1}f(x^2)dx+b\int_{0}^{1}g(x^2)dx=a\phi(f)+b\phi(g)$.
My confusion is on boundedness and its norm.

If you change variables, your integral will look like
$$\int_0^1 f(x) (d x/(2 \sqrt{x})).$$
This is an integral with respect to a finite measure, so is bounded. To find the norm, use Holder's inequality.
