For which real numbers \(\alpha,\beta\) does \[ \|f\|_2\le\|f\|_{4/3}^{\alpha}\|f\|_4^{\beta} \] hold for all measurable \(f\) for which the right-hand side is finite?
Solution
The inequality holds exactly for \[ \boxed{\alpha=\beta=\frac12}. \]
1. Necessity.
Apply the inequality to \(cf\) with \(c>0\). Homogeneity forces \[ 1=\alpha+\beta. \] Next take \(f=\mathbf1_E\) for measurable sets \(E\) of arbitrary finite positive measure \(m(E)=m\). Then \[ \|f\|_2=m^{1/2}, \qquad \|f\|_{4/3}=m^{3/4}, \qquad \|f\|_4=m^{1/4}. \] Thus \[ m^{1/2}\le m^{3\alpha/4+\beta/4} \qquad\text{for every }m>0. \] For this to hold both as \(m\downarrow0\) and \(m\to\infty\), the exponents must agree: \[ \frac12=\frac{3\alpha+\beta}{4}. \] Together with \(\alpha+\beta=1\), this gives \[ \alpha=\beta=\frac12. \]
2. Sufficiency.
By Hölder’s inequality with exponents \(4/3\) and \(4\), \[ \int |f|^2 =\int |f|\,|f| \le\left(\int|f|^{4/3}\right)^{3/4}\left(\int|f|^4\right)^{1/4}. \] Taking square roots gives \[ \|f\|_2\le\|f\|_{4/3}^{1/2}\|f\|_4^{1/2}. \] Thus the claimed pair is both necessary and sufficient.