Problem Statement.
I am working on the following exercise:
Prove that if $\mu$ is a measure, $p\in(1,\infty)$ and $f,g\in L^p(\mu)$ are such that $$||f||_p=||g||_p=\left|\left|\frac{f+g}{2}\right|\right|_p,$$ then $f=g$.
Notation.
For a measure space $(X,\mathcal{S},\mu)$ and $p\in(0,\infty]$ (where $\mathbf{F}$ is taken to be either $\mathbf{R}$ or $\mathbf{C}$):
$L^p(\mu)=\{\tilde{f}: f\in\mathcal{L}^p(\mu)\}$
$\mathcal{L}^p(\mu)=\{f:X\rightarrow\mathbf{F}: f \text{ is } \mathcal{S}\text{-measurable and } \int f\,d\mu<\infty\}$
$\tilde{f}=\{f+z: z\in\mathcal{Z}(\mu)\}$ where $\mathcal{Z}(\mu)$ is the set of $\mathcal{S}$-measurable functions from $X$ to $\mathbf{F}$ that equal $0$ almost everywhere.
My Thoughts.
I am a bit confused as to how this is even an exercise, and here is why. We define $$||\tilde{f}||_p=||f||_p$$ for $p\in (0,\infty].$ Thus even though the $f$ and $g$ given are actually $\tilde{f}$ and $\tilde{g}$, it doesn't matter since $||\tilde{f}||_p=||f||_p$ and $||\tilde{g}||_p=||g||_p$, thus implying that $||f||_p=||g||_p$. But if this is the case then $$\left(\int |f|^p\,d\mu \right)^{1/p}=\left(\int |g|^p\,d\mu \right)^{1/p}.$$ How could that possibly be the case if $f\neq g$? I would say I'm done at this point, but clearly I am missing something. I didn't even use the last equality, i.e. $$\left|\left|\frac{f+g}{2}\right|\right|_p.$$ So what am I missing here? There must be something I am misunderstanding!
Thanks in advance for your help!
The p-norm $f \mapsto \|f\|_p$ is strictly convex (edit : as a norm on a vector space, not as a function), due to Minkowski inequality.
I will note $\| \cdot \|=\| \cdot \|_p$ for convenience.
If $f \ne \lambda g$ on a set of positive measure for all $\lambda >0$, then for all $t \in (0,1)$ :
$$\|t f +(1-t)g\|<t\|f\| +(1-t)\|g\|.$$
But, having $\|f\|=\|g\|$ being equal to $\left\|\frac{f+g}2 \right\|$ is impossible. Indeed, putting $t = 1/2$ yields :
$$\left\|\frac{f+g}2 \right\|<\frac 1 2 (\| f \| + \|g \|) =\|f\|=\|g\|.$$
Then $f= \lambda g$, but since $\|f\|=|\lambda| \|g\|$ and $\lambda >0$, then $\lambda = 1$.
So we must have $f=g$ a.e. .