I'm reading a paper which uses the Besov space $B = B^{\infty, \infty}_{-1}$ defined for $f: \mathbb{R}^3 \to \mathbb{C}$ by
$$ \| f \|_{B} = \sup_{t > 0} t^{1/2} \| e^{t\Delta} f(x)\|_{L^\infty_x(\mathbb{R}^3)}.$$
Here $e^{t\Delta}$ is the heat flow, so if $ f(x) = \int_{\mathbb{R}^3} e^{i k \cdot x} \hat{f}(k) \; dk, $ then $$ e^{t\Delta} f(x) = \int_{\mathbb{R}^3} e^{i k \cdot x - t|k|^2} \hat{f}(k) \; dk. $$
In the paper, the authors seem to claim that $$ \| f\|_{B} \leq C \|f\|_{L^\infty_x(\mathbb{R}^3)}.$$
My question is, is this inequality true in general, or do I need some special property of the function $f$? The authors use the inequality without comment, but it's possible they're relying on some feature of the [complicated] functions they're bounding.
It's simple to see from the Fourier representation above that if $ \operatorname{supp} \hat{f} $ is contained in $\{ k : |k| > \epsilon >0 \}$, then we have
$$ \| f\|_{B} \leq \frac{C}\epsilon \|f\|_{L^\infty_x(\mathbb{R}^3)},$$ since $$\sup_{t > 0} \sup_{|k| > \epsilon} t^{1/2} e^{-|k|^2 t} = C/\epsilon.$$
I don't see a way to deal with the low frequencies though. Any insight/counterexamples/intuition would be appreciated.
Consider the function $f \in L^\infty(\mathbb{R}^n)$ given by $f(x) =1$ for all $x$. Since the heat kernel has unit mass for all $t >0$ we see that $e^{t \Delta} f(x) =1$ for $t >0$ and $x \in \mathbb{R}^n$, and so $\Vert e^{t \Delta} f\Vert_{L^\infty} = 1$. Thus $$ \Vert f \Vert_B = \sup_{t >0} t^{1/2} \Vert e^{t \Delta} f\Vert_{L^\infty} = \infty. $$ Given this, we find that the stated inequality $\Vert f\Vert_B \le C \Vert f \Vert_{L^\infty}$ is false for general $f \in L^\infty$.