Show $\|f\|_p\leq\|f\|_q\mu(X)^{\frac{1}{p}-\frac{1}{q}}$ whenever $0<p\leq q\leq \infty$ using log-convexity of $L^p$ norms.

Question

Assume that the measure space is finite for this to make sense. Also, we know that $L^p$ spaces satisfy log convexity, that is - $$\|f\|_r \leq \|f\|_p^\theta \|f\|_q^{1-\theta}$$ where $\frac{1}{r}=\frac{\theta}{p} +\frac{1-\theta}{q}$. The text which I am following says 'Indeed this is trivial when $q=\infty$, and the general case then follows by convexity'. I understand that it is true when $q=\infty$, however I am unable to use that it is true for $q=\infty$ for proving the general case. I have found a proof which uses log-convexity and the fact that $\lim_{p\rightarrow 0}\|f\|_p^p=\mu(\text{supp}f)$. Is there some way that I can do this by using that it is true for $q=\infty$?

Answer 1

Since $\frac{q}p>1$ then the map $\phi: x\mapsto x^{\frac{q}p}$ is convex and therefore by Jenssens inequality(which is convexity) we have

$$\phi\left(\frac{1}{\mu(X)}\int_X|f|^p d\mu\right)\le \frac{1}{\mu(X)}\int_X\phi\circ|f |^pd\mu $$ that is $$\left(\frac{1}{\mu(X)}\int_X|f|^p d\mu\right)^{q/p}\le \frac{1}{\mu(X)}\int_X|f |^qd\mu $$

that is $$\|f\|_p\leq\|f\|_q\mu(X)^{\frac{1}{p}-\frac{1}{q}}$$

Answer 2

By Cauchy-Schwarz, for $q\ge p$, $$ \begin{align} \|f\|_p &=\left(\int_X|f(x)|^p\,\mathrm{d}x\right)^{\frac1p}\\ &=\left(\int_X|f(x)|^p\cdot1\,\mathrm{d}x\right)^{\frac1p}\\ &\le\left(\int_X|f(x)|^q\,\mathrm{d}x\right)^{\frac1q} \left(\int_X1^{\frac q{q-p}}\,\mathrm{d}x\right)^{\frac1p-\frac1q}\\[9pt] &=\|f\|_q\,\mu(X)^{\frac1p-\frac1q} \end{align} $$