Heat semigroup on $C_b(\mathbb R)$

Question

Let $(X,\|\cdot\|)\in \{(L^2(\mathbb R),\|\cdot\|_{L^2}),(L^\infty(\mathbb R),\|\cdot\|_{L^\infty}),(C_b(\mathbb R),\|\cdot\|_{\infty})\}$

I have a question regarding the heat semigroup $$T_tf:=(\varphi_t\ast f)$$ for $f\in X$, $t>0$ and $$\varphi_t\colon \mathbb R\to \mathbb R,\; x\mapsto \frac1{\sqrt{4\pi t}}\exp\left(-\frac{y^2}{4t}\right),\; t>0.$$ Since $(\varphi_t)_{t>0}$ is a mollifier, it is relatively easy to calculate that $(T_t)_{t\geq0}$ forms a strongly continuous semigroup on $X=L^2(\mathbb R)$. However, it is not strongly continuous on $X=L^\infty(\mathbb R)$. To do this, you can choose $f=\mathbb1_{[0,\infty)}\in L^\infty(\mathbb R)$ and show that $$\|T_tf-f\|_{L^\infty}\not\to 0$$ applies. I have now asked myself what happens for $X=C_b(\mathbb R)$. The idea must be similar to $X=L^\infty(\mathbb R)$, but the $f$ chosen there is not continuous. The obvious approach would now be to smooth $f$, unfortunately linear interpolation does not lead to the desired result. Hence my question, is there a "simple" function to disprove the strong continuity of $(T_t)_{t\geq0}$ on $C_b(\mathbb R)$?

Answer 1

$f(x) = \sin x^2$ is continuous and bounded, but not uniformly continuous. For such functions, in general, translation is not $L^\infty$ continuous, and thus there might be some problems with convolution.

I didn't compute the required integral, but looking at a couple of plots, it seems that $T_t f$ is always vanishing at infinity. Therefore it cannot converge in $L^\infty$ to $f$.

Answer 2

Maybe $ C(\mathbb{R}) $ is not a suitable space for studying the heat group because the functions in this space can grow arbitrarily large at infinity. For example, if you take $ f = \mathrm{e}^{x^4} $, which belongs to $ C(\mathbb{R}) $, $ T_t f $ is not well-defined because the integral $ \phi_t * f $ does not converge.