Let $f \in H^{1}(\mathbb{R}^{d})$ and $t > 0$. I want to prove that: $$\mathcal{F}({e^{t\Delta}f})(k) = e^{-t|2\pi k|^{2}}\mathcal{F}(f)(k),$$ where $\Delta$ denotes the Laplacian operator on $\mathbb{R}^{d}$ and $\mathcal{F}$ denotes the Fourier transform: $$(\mathcal{F}f)(k) = \int_{\mathbb{R}^{d}}e^{-2\pi i \langle k,x\rangle}f(x)dx.$$
I know the following result from functional analysis: let $\mathscr{H}_{1}$ and $\mathscr{H}_{2}$ be Hilbert spaces and $T: \mathscr{H}_{1} \to \mathscr{H}_{1}$ and $S: \mathscr{H}_{2} \to \mathscr{H}_{2}$ be two self-adjoint (possibly unbounded) which are unitarily equivalent, that is, $T = U^{-1}SU$ for some unitary operator $U: \mathscr{H}_{1} \to \mathscr{H}_{2}$. Then, for every continuous and bounded function $f: \mathbb{R} \to \mathbb{C}$, one has $f(T) = U^{-1}f(S)U$.
Can I use this result to prove the formula above? My idea is: let $f(x) = e^{-x}$ and consider $T = -t\Delta$ and $U = \mathcal{F}$, which is an unitary map. The point that is not clear to me is: $f(x) = e^{-x}$ is not bounded on $\mathbb{R}$, but it is bounded on $[0,\infty)$, which is the spectrum of $-t \Delta$. Can I use the formula in this case? In other words, is it enough that $f$ is bounded and continuous on the spectrum of $T$?
Let $E$ be the spectral measure of $- \Delta$.
First of all we do not need a bounded function to use the spectral theorem, it only needs to be measurable.
To answer the question in the last paragraph: The spectral measure is supported in the spectrum, meaning that $E(\sigma (- \Delta)) = I$ from which it follows that $E(\mathbb{R}\setminus \sigma( - \Delta)) = 0$. Therefore the values outside of the spectrum of the measurable function $f: \mathbb{R} \to \mathbb{R}$ for which we want to define $f(-\Delta)$ do not influence $f(-\Delta)$. Meaning that we can redefine $f$ outside the spectrum as we please (for example set it to $0$). Therefore $f(-\Delta)$ will be continuous if $f$ is bounded on the spectrum.
In this case we have that $- \Delta$ is unitarily equivalent via the Fourier transform $\mathcal{F}$ to the operator $M_g$ that is given by the multiplication with the function $g$ defined by $g(k)= |2\pi k|^2 $. Let $F$ be the spectral measure of $M$. Per this post we know that $E = \mathcal{F}^* F \mathcal{F}$. It can be shown that for a multiplication operator $f(M_g)= M_{f \circ g}$ and therefore $$f(- \Delta ) =\mathcal{F}^* f(M_g)\mathcal{F} = \mathcal{F}^* M_{f \circ g} \mathcal{F}.$$ The desired formula is obtained by setting (for a fixed $t>0$) $f (x)= \exp (-t x)$ for $ x\in \mathbb{R}$. The formula also holds for all $L^2$ functions, since $\exp (t \Delta)$ is bounded (and not just on $H^1$).
Most of this post works generally and not just for the specific operator under consideration here.