Why $|| f^{(k)} ||_{BMO} \to || f||_{BMO}$?

Question

We also deduce that if $f\in BMO$ is real-valued, and $f^{(k)}$ is the truncation of $f$ defined by $f^{(k)}(x)=f(x)$ if $\mid f(x)\mid\leq k;f^{(k)}(x)=k$ if $f(x)>k$; and $f^{(k)}(x)=-k$, if $f(x)<-k$, then $\{f^{(k)}\}$ is a sequence of bounded BMO functions so that $\mid f^{(k)}\mid\leq \mid f\mid $ for all $k,f^{(k)}\to f$ for a.e $x$ as $k\to\infty$, and hence $|| f^{(k)} ||_{BMO} \to || f||_{BMO}$ as $k\to \infty$. If $f$ is complex-valued, one may apply this to both the real and imaginary parts of $f$.

1. Why is $f^{(k)}\to f$ almost everywhere, not everywhere? Does this suggest that for BMO functions and in general harmonic analysis, we allow $f$ to have the value $\infty$?

2. How does that make $|| f^{(k)} ||_{BMO} \to || f||_{BMO}$? The book suggests that we need to use the fact that if $f$ and $g$ belong to $BMO$ then so do $min(f,g)$ and $max(f,g)$.

3. Can you please briefly how I can generalize this to complex-valued case?

Thank you very much.

Answer 1

I just manage to get one-sided estimate for 2).

First of all,

1. Yes, since Lebesgue measurable requires defined a.e. and so if $f(x)=\infty$ on a set of measure zero, then the truncation valid a.e. only

2. Note that with Lebesgue dominated convergence theorem, one has $\text{Avg}_{B}f^{(k)}\rightarrow\text{Avg}_{B}f$ and hence $|f^{(k)}(x)-\text{Avg}_{B}f^{(k)}|\rightarrow|f(x)-\text{Avg}_{B}f|$ a.e.

Fatou lemma gives \begin{align*} \dfrac{1}{|B|}\int_{B}|f(x)-\text{Avg}_{B}f|&\leq\liminf_{k\rightarrow\infty}\dfrac{1}{|B|}\int_{B}|f^{(k)}(x)-\text{Avg}_{B}f^{(k)}|\\ &\leq\liminf_{k\rightarrow\infty}\|f^{(k)}\|_{\text{BMO}}, \end{align*} so $\|f\|_{\text{BMO}}\leq\liminf_{k\rightarrow\infty}\|f^{(k)}\|_{\text{BMO}}$.

3. Note that with $f=\text{Re}f+i\text{Im}f$, then \begin{align*} \int_{B}|f(x)-\text{Avg}_{B}f|&=\int_{B}|\text{Re}f+\text{Im}f-\text{Avg}_{B}\text{Re}f-\text{Avg}_{B}\text{Im}f|\\ &\leq\int_{B}|\text{Re}f-\text{Avg}_{B}\text{Re}f|+\int_{B}|\text{Im}f-\text{Avg}_{B}\text{Im}f|, \end{align*} so it is still the matter of real-valued function taken into account.