Is the map $x\mapsto\|x\|_p^p$, defined in $L^p[0,1]$ ($1<p<\infty$), strictly convex?

Question

Is the map $x\mapsto\|x\|_p^p$, defined in $L^p[0,1]$ (for $1<p<\infty$), strictly convex?

Maybe there is a classic inequality that can give the conclusion easily. Let me know if this is the case.

Edit: In the case it is not strictly convex, if we restrict the domain of $x\mapsto\|x\|_p^p$ to be the unit ball in $L^p[0,1]$, the map is strictly convex?

Answer 1

We have for all $x,y$ and $\lambda\geq 0$, by the triangle (Minkowski) inequality: $$|\!|\lambda x+(1-\lambda) y)|\!|_p^p\leq\left(\lambda|\!|x|\!|_p+(1-\lambda)|\!|y|\!|_p\right)^p$$ and the result follows from the strict convexity of the function $t\to t^p$ for $p>1$ and $t\geq 0$.

Answer 2

Minkowski's inequality works: $$\begin{align}||\lambda g+(1-\lambda)f||_p=\left(\int_X (|\lambda g+(1-\lambda) f|)^p \text{d}\mu \right)^{\frac 1p} &\le \left(\int_X (\lambda |g|)^p \text{d}\mu \right)^{\frac 1p}+\left(\int_X ((1-\lambda) |f|)^p \text{d}\mu \right)^{\frac 1p}\\ &=\lambda ||g||_p+(1-\lambda)||f||_p \end{align}$$