Can we show that $\int\left|1_Bf-\int_Bf\:{\rm d}\lambda\right|^2\:{\rm d}\lambda\le c\int\left|f-\int f\:{\rm d}\lambda\right|^2\:{\rm d}\lambda$?

Question

Let $(E,\mathcal E,\lambda)$ be a $\sigma$-finite measure space, $f:E\to[0,\infty)^3$ be $\mathcal E$-measurable with $$\int|f|^2\:{\rm d}\lambda<\infty,\tag1$$ $k\in\mathbb N$ and $B_1,\ldots,B_k\in\mathcal E$ be disjoint with $$E=\bigcup_{i=1}^kB_i.\tag2$$

Are we able to derive a bound $$\max_{1\le i\le k}\int\left|1_{B_i}f-\int_{B_i}f\:{\rm d}\lambda\right|^2\:{\rm d}\lambda\le c\int\left|f-\int f\:{\rm d}\lambda\right|^2\:{\rm d}\lambda\tag3$$ for some $c\ge0$?

For example, if $\lambda$ is finite, $f=1_A$ (so, assuming it is scalar-valued) for some $A\in\mathcal E$ and $k=1$, $(3)$ reduces to \begin{equation}\begin{split}&\lambda(A\cap B)-2\left|\lambda(A\cap B)\right|^2+\lambda(E)\left|\lambda(A\cap B)\right|^2\\&\;\;\;\;\;\;\;\;\;\;\;\;\le c\left(\lambda(A)-2|\lambda(A)|^2+\lambda(E)|\lambda(A)|^2\right).\end{split}\tag3\end{equation}

Answer 1

Take $(E,\mathcal{E})=([0,1],\mathcal{B}_{[0,1]})$ equipped with the Lebesgue measure $\lambda$, and $f\equiv 1$. Then $$ \int\left|f-\int f\,d\lambda\right|^2\,d\lambda=0. $$ However, for $B_1=[0,1/2]$ and $B_2=(1/2,1]$, $$ \max_{1\le i\le 2}\int\left|f1_{B_i}-\int_{B_i} f\,d\lambda\right|^2\,d\lambda=1/4. $$