Let $f \in L^p (\Omega)$ and $\tilde{f}=\frac{1}{|\Omega|}\int_{\Omega}f$, where $\Omega \subset R^N$ is open and bounded, I want to know if the following statement is true or false
$\| f- \tilde{f} \|_p \leq \| f \|_{p}$
I have tried with some functions and I think it is true. If the mean is zero there is nothing to do, in otherwise if $\tilde{f}$ is different to zero, I believe that the function $f-\tilde{f}$ will be closer to the reference axes so its integral will have a lower value. but I have no idea how to prove or refute the statement, if you could give me an idea. I have this
Lemm1. There exist C independent f such that $\| f- \tilde{f} \|_p \leq C \| f \|_{p}$
Proof: $\| f- \tilde{f} \|_p \leq \|f\|_p+\| \tilde{f} \|_p=\|f\|_p+|\tilde{f}| |\Omega|^{\frac{1}{p}}$ $$= \|f\|_p+|\Omega|^{\frac{1}{p}-1}|\int_{\Omega}f| $$ $$\leq \|f\|_p+|\Omega|^{\frac{1}{p}-1}\|f\|_1$$ Since $L^p \subset L^1$ we obtain the statement.
Thank you. I'm sorry for not giving much in the solution. excuse my english.
Sorry initially I asked the contrary statement this was all my fault
I have this conjecture $\| f- \tilde{f} \|_p = \inf_{k \in K}\| f +k\|_{p}$ False. See comments Where K is the set of all constants.
Consider $L^1[0, 1]$ and $f(x) = 1+x$, then we see that \begin{align} \tilde f = \|f\|_1=\int^1_0 1+x\ dx = \frac{3}{2} \end{align} which means \begin{align} \|f-\tilde f\|_1 = \int^1_0 \left|x-\frac{1}{2}\right|\ dx = \frac{1}{4}. \end{align} Hence the stated inequality doesn't hold.
However, a better question would be: Does there exists $C>0$ depending only on $\Omega=[0, 1]$ such that \begin{align} \|f\|_1 \leq C\|f-\tilde f\|_1? \end{align}