Exercise:
For which $p\in [1,\infty]$ is $L^p([0,1])$ strictly convex?
Solution: For strict convexity we have two equivalent definition:
- If $x\neq 0\neq y$ and $\Vert x+y\Vert=\Vert x\Vert+\Vert y\Vert$ then $x=\lambda y$ for some $\lambda$.
- If $x\neq y$ with $\Vert y\Vert=\Vert x\Vert=1$ then $\Vert\frac{x+y}{2}\Vert<1$
So I found that $L^p([0,1])$ is strictly convex for $p=2$: \begin{equation} \int_0^1(\frac{f(t)+g(t)}{2})^2dt=0.5\int_0^1(f(t)^2+g(t)^2+2f(t)g(t))dt\leq \frac{\Vert f\Vert^2_2+\Vert g\Vert^2_2}{2}<1 \end{equation}
Question: Are there other $p$ for which $L^p([0,1])$ is strictly convex?
