If $r

Question

Here, $||f||_p$ denotes the $L^p$-norm. I have so far managed to prove that if $\phi (p) =||f||_p^p$ then $\phi $ is convex and that $\phi(p)\le \max (\phi(s),\phi(r))$. Now all that I need to prove is $\phi(p)^{\frac 1p}\le \max (\phi(s)^{\frac 1s},\phi(r)^{\frac 1r})$ but I have no idea how to do it. I thought of using the Holder inequality with $\log \phi$ but I do not have clear conjugate exponents. I tried proving it by cases but my attempts have guided me nowhere. Where do I start? $$ $$Edit
An answer was posted but was for unknown reasons deleted. I am writing it down here in case it comes in handy for future readers of this post and because I found it really nice:
Since $r<p<s\Rightarrow \frac 1s <\frac 1p<\frac 1r \Rightarrow \exists \lambda \in(0,1) \ \text{s.t.} \ \ \frac 1p=\frac \lambda s+\frac{1-\lambda}{r} \Rightarrow 1=\frac {p\lambda} s+\frac{p(1-\lambda)}{r} $. In this case, define $T= \frac{s}{p\lambda}$ and $Q=\frac{r}{p(1-\lambda)}$, then $Q$ and $T$ are conjugate exponents and $$\begin{align} ||f||_p=\left\{\int_X |f|^p \text{d}\mu\right\}^{\frac 1p} &=\left\{\int_X |f|^{\lambda p}|f|^{(1-\lambda )p}\text{d}\mu\right\}^{\frac 1p}\\&\le \left\{\left(\int_X |f|^{\lambda pT}\text{d}\mu\right)^{\frac 1T}\left(\int_X |f|^{(1-\lambda) pQ}\text{d}\mu\right)^{\frac 1Q}\right\}^{\frac 1p}\tag{Hölder's inequality} \\ &=\left\{\left(\int_X |f|^{\lambda p\frac {s}{p\lambda}}\text{d}\mu\right)^{\frac {p\lambda}{s}}\left(\int_X |f|^{(1-\lambda) p\frac {r}{p(1-\lambda)}}\text{d}\mu\right)^{\frac {p(1-\lambda)}{r}}\right\}^{\frac 1p} \\ &=\left\{\left(\int_X |f|^{s}\text{d}\mu\right)^{\frac {p\lambda}{s}}\left(\int_X |f|^{ r}\text{d}\mu\right)^{\frac {p(1-\lambda)}{r}}\right\}^{\frac 1p}\\ &=\left(\int_X |f|^{s}\text{d}\mu\right)^{\frac {\lambda}{s}}\left(\int_X |f|^{ r}\text{d}\mu\right)^{\frac {(1-\lambda)}{r}} \\ &=||f||_s^{\lambda}||f||_r^{1-\lambda}\end{align}$$ Now suppose $\max \{||f||_s,||f||_r\} = ||f||_s$ then $||f||_r \le ||f||_s \Rightarrow ||f||_r^{1-\lambda} \le ||f||_s^{1-\lambda} \Rightarrow ||f||_r^{1-\lambda}||f||_s^{\lambda} \le ||f||_s^{1}$, if $\max \{||f||_s,||f||_r\} = ||f||_r$ then in the above argument map $s \mapsto r$, $r \mapsto s$, $\lambda \mapsto 1-\lambda$, and $1-\lambda \mapsto \lambda$.

Answer 1

Using Hölder's inequality is a good idea. There is $t\in (0,1)$ such that $$ \frac{1}{p} = \frac{1-t}{r} + \frac{t}{s} $$ (any number between two others is their convex combination). Write $|f|^p = |f|^{p(1-t)} |f|^{pt}$ and apply the inequality as $$ \int|f|^{p(1-t)} |f|^{pt} \le \left(\int |f|^r\right)^{p(1-t)/r} \left(\int |f|^s\right)^{pt/s} $$ Raising both sides to power $1/p$ yields $$ \|f\|_p \le \|f\|_r^{1-t} \|f\|_s^t \le \max(\|f\|_r,\|f\|_s) $$