Is multiplication by bounded, continuous functions bounded on $BMO$?

Question

I know that if $f\in L^\infty$ then the multiplication operator $\varphi\mapsto\varphi f$ is not bounded on $BMO$, see eg. here. The obvious counterexample is something like $f=1_{x\geq0}$ and $\varphi=\log|x|$, since the $BMO$ness of $\log|x|$ relies on the function diverging at zero from the left and right. But what happens if we assume $f$ is also continuous to rule out counterexamples like this one? Can we now say $\|\varphi f\|_{BMO}\leq\|\varphi\|_{BMO}\|f\|_{L^\infty}$?

Answer 1

A modification of the counterexample given in the linked thread yields a counterexample for this case. Approximate $\mathbb{1}_{x > 0}$ by continuous functions $f_n$. Have $f_n(x) = 1$ for $x > 0$ and $f_n(x) = 0$ for $x < -1/n$ and $f_n(x) \in [0,1]$ always. Then set $g_n(x) = f_n(x) \log \lvert {x} \rvert$. Let $Q = Q_n = [-2/n, 2/n]$. We have $\lvert g(x) - (g_n)_Q\rvert $ is larger than the half the average of $\lvert \log \lvert y \rvert\rvert$ over $[-1/n, 1/n]$ for $x \in [-2/n, -1/n]$. Thus $$ \| g_n \|_{\text{BMO}} \geq \left( \frac{n}{4} \cdot \frac{1}{n} \cdot (\frac{1}{2} \lvert \log 1/n \rvert )^2 \right)^{1/2} = \frac{1}{4} \lvert \log n \rvert $$ diverges, while $\| \log \lvert x \rvert \|_{\text{BMO}} \| f_n \|_\infty$ is bounded.