Htykuut

Problem Statement.

I am working on the following exercise:

Prove that if $mu$ is a measure, $pin(1,infty)$ and $f,gin
L^p(mu)$ 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 $pin(0,infty]$ (where $mathbf{F}$ is taken to be either $mathbf{R}$ or $mathbf{C}$):

1. $L^p(mu)={tilde{f}: finmathcal{L}^p(mu)}$




2. $mathcal{L}^p(mu)={f:Xrightarrowmathbf{F}: f text{ is } mathcal{S}text{-measurable and } int f,dmu<infty}$




3. $tilde{f}={f+z: zinmathcal{Z}(mu)}$ 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 $pin (0,infty].$ 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 $$left(int |f|^p,dmu right)^{1/p}=left(int |g|^p,dmu right)^{1/p}.$$ How could that possibly be the case if $fneq g$? 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. .