If $f : [0,\infty) \to [0,\infty)$ is an integrable function, then what is the best possible constant $k$, for which the following ineqality holds:
$$\int_0^{\infty}f(x)dx \leq k\left(\int_0^{\infty}\sqrt{x}f(x)dx\right)^{1/2}\cdot\left(\int_0^{\infty}f^{2}(x)dx\right)^{1/4}$$
For example, $k=2$ is a bound, since, $$\int_0^{\infty}f(x)dx = \int_0^{y}f(x)dx + \int_y^{\infty}f(x)dx < \sqrt{y}\left(\int_0^{\infty}f^2(x)dx\right)^{1/2} + \frac{1}{\sqrt{y}}\int_0^{\infty}\sqrt{x}f(x)dx$$
and setting, $ y = \dfrac{\displaystyle \int_0^{\infty}\sqrt{x}f(x)dx}{\left(\displaystyle\int_0^{\infty} f^2(x)dx\right)^{1/2}}$, establishes the inequality for the case $k=2$, but it is not strict.
How to improve the bound $k$ and also determine the function where equality holds for the best constant $k$ ?
Thank you.
Suppose $$ \int_0^\infty\sqrt{x}f(x)\,\mathrm{d}x=A\tag{1} $$ and $$ \int_0^\infty f^2(x)\,\mathrm{d}x=B\tag{2} $$ Then, letting $g(x)^2=A^{1/2}B^{-3/4}f(AB^{-1/2}x)$, we get $$ \int_0^\infty\sqrt{x}g(x)^2\,\mathrm{d}x=\int_0^\infty g(x)^4\,\mathrm{d}x=1\tag{3} $$ and $$ \int_0^\infty f(x)\,\mathrm{d}x=A^{1/2}B^{1/4}\int_0^\infty g(x)^2\,\mathrm{d}x\tag{4} $$ Thus, maximizing $\int_0^\infty g(x)^2\,\mathrm{d}x$ given $(3)$ yields the best constant for the given inequlity.
The variation of $(4)$ is stationary when $$ \int_0^\infty g(x)\,\delta g(x)\,\mathrm{d}x=0\tag{5} $$ Equations $(3)$ imply $$ \int_0^\infty\sqrt{x}g(x)\,\delta g(x)\,\mathrm{d}x=\int_0^\infty g(x)^3\,\delta g(x)\,\mathrm{d}x=0\tag{6} $$ To insure $(5)$ holds whenever $(6)$ does, we must have that $g(x)^2=0$ or $g(x)^2=a-b\sqrt{x}$.
Thus, consider $g(x)^2=a-b\sqrt{x}$ on $\left[0,\frac{a^2}{b^2}\right]$ and $g(x)^2=0$ for $x\gt\frac{a^2}{b^2}$. $$ \begin{align} \int_0^{a^2/b^2}\sqrt{x}g(x)^2\,\mathrm{d}x &=\int_0^{a^2/b^2}\left(a\sqrt{x}-bx\right)\,\mathrm{d}x\\ &=\frac{a^4}{6b^3}\tag{7} \end{align} $$ $$ \begin{align} \int_0^{a^2/b^2}g(x)^4\,\mathrm{d}x &=\int_0^{a^2/b^2}\left(a^2-2ab\sqrt{x}+b^2x\right)\,\mathrm{d}x\\ &=\frac{a^4}{6b^2}\tag{8} \end{align} $$ $$ \begin{align} \int_0^{a^2/b^2}g(x)^2\,\mathrm{d}x &=\int_0^{a^2/b^2}\left(a-b\sqrt{x}\right)\,\mathrm{d}x\\ &=\frac{a^3}{3b^2}\tag{9} \end{align} $$
Solving $(7)$ and $(8)$ given $(3)$ yields $$ (a,b)=\left(6^{1/4},1\right)\tag{10} $$ Plugging $(10)$ into $(9)$ yields an optimal constant of $$ \left(\frac83\right)^{1/4}\doteq1.277886208492545\tag{11} $$ attained by the function $$ \left(6^{1/4}-\sqrt{x}\right)\left[0\le x\le6^{1/2}\right]\tag{12} $$ where $\left[\dots\vphantom{6^{1/2}}\right]$ are Iverson brackets.
The Slick Answer
Suppose $f(x)\ge0$ and $$ \int_0^\infty\sqrt{x}f(x)\,\mathrm{d}x=\int_0^\infty f(x)^2\,\mathrm{d}x=1\tag{13} $$ Then letting $u(x)=\left(6^{1/4}-\sqrt{x}\right)\left[0\le x\le6^{1/2}\right]$ $$ \begin{align} 0 &\le\int_0^\infty(f(x)-u(x))^2\,\mathrm{d}x\\ &=\int_0^\infty\left(f(x)^2-2f(x)u(x)+u(x)^2\right)\,\mathrm{d}x\tag{14} \end{align} $$ Applying $(13)$ to $(14)$, dividing by $2$, and rearranging, we get $$ \begin{align} 1 &\ge\int_0^\infty f(x)u(x)\,\mathrm{d}x\\ &\ge\int_0^\infty f(x)\left(6^{1/4}-\sqrt{x}\right)\,\mathrm{d}x\\ &=6^{1/4}\int_0^\infty f(x)\,\mathrm{d}x-1\tag{15} \end{align} $$ Therefore, $$ \int_0^\infty f(x)\,\mathrm{d}x\le\left(\frac83\right)^{1/4}\tag{16} $$