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Stationary Solutions to the Stochastic Burgers Equation on the Line


We consider invariant measures for the stochastic Burgers equation on \({\mathbb {R}}\), forced by the derivative of a spacetime-homogeneous Gaussian noise that is white in time and smooth in space. An invariant measure is indecomposable, or extremal, if it cannot be represented as a convex combination of other invariant measures. We show that for each \(a\in {\mathbb {R}}\), there is a unique indecomposable law of a spacetime-stationary solution with mean a, in a suitable function space. We also show that solutions starting from spatially-decaying perturbations of mean-a periodic functions converge in law to the extremal space-time stationary solution with mean a as time goes to infinity.

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We are happy to thank Yuri Bakhtin and Konstantin Khanin for generous explanations of their work, and Kevin Yang for productive discussions. We are also especially grateful to Yu Gu for pointing out a strengthening of Proposition 5.2 that led to an improvement of our main result over the initial version of the paper. AD was partially supported by an NSF Graduate Research Fellowship under grant DGE-1147470, CG by the Fannie and John Hertz Foundation and NSF grant DGE-1656518, and LR by NSF grants DMS-1613603 and DMS-1910023, and ONR grant N00014-17-1-2145.

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Correspondence to Alexander Dunlap.

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Proof of Proposition 1.4

We now prove Proposition 1.4. The proof is elementary and independent of the rest of the paper. Let \(\varepsilon >0\). For each \(j=1,\ldots ,N\), define \(w_{j}\in L^{\infty }({\mathbb {R}})\) by

$$\begin{aligned} w_{j}(x)&=\sup _{|y|\ge |x|}|v_{\mathrm {z},j}(y)|, \end{aligned}$$

so \(|v_{\mathrm {z},j}(x)|\le w_{j}(x)\) for all \(x\in {\mathbb {R}}\). Then, \(w_{j}(x)\) is decreasing in |x|, and

$$\begin{aligned} \lim _{|x|\rightarrow \infty }w_{j}(x)=0. \end{aligned}$$

Therefore, we can find a \(K\in {\mathbb {N}}\) so large that

$$\begin{aligned} \frac{1}{KL}\max _{j=1}^{N}\Vert v_{\mathrm {int},j}\Vert _{L^{1}({\mathbb {R}})}&<\varepsilon /2,&\frac{1}{KL}\max _{j=1}^{N}\int _{0}^{KL}w_{j}(x)\,{d}x&<\varepsilon /2. \end{aligned}$$

Let us define

$$\begin{aligned} v_{\pm ,j}(x)&=v_{\mathrm {per},j}(x)\pm w_{j}(x)\pm \sup _{m\in {\mathbb {Z}}}|v_{\mathrm {int},j}(x+mKL)| \end{aligned}$$

and \({\mathbf {v}}_{-}=(v_{-,1},\ldots ,v_{-,N})\) and \({\mathbf {v}}_{+}=(v_{+,1},\ldots ,v_{+,N})\), so \({\mathbf {v}}_{-},{\mathbf {v}}_{+}\in L^{\infty }({\mathbb {R}})^{N}\) and \({\mathbf {v}}_{-}\preceq {\mathbf {v}}\preceq {\mathbf {v}}_{+}\). Then we have

$$\begin{aligned}&\frac{1}{KL}\int _{0}^{KL}{\mathbf {v}}_{-}(x)\,{d}x-{\mathbf {a}} \\&\quad =\frac{1}{KL}\int _{0}^{KL}[{\mathbf {v}}_{-}(x) -{\mathbf {v}}_{\mathrm {per}}(x)]\,{d}x \\&\quad =\frac{1}{KL}\int _{0}^{KL}\left[ -w_{j}(x)+\inf _{i\in {\mathbb {Z}}} v_{\mathrm {int},j}(x+iKL)\right] \,{d}x\\&\quad \succeq -\frac{\varepsilon }{2}(1,\ldots ,1)-\frac{1}{KL}\max _{j=1}^{N}\int _{0}^{KL}\sum _{i\in {\mathbb {Z}}}\left| v_{\mathrm {int},j}(x+iKL)\right| \,{d}x \\&\quad =-\frac{\varepsilon }{2}(1,\ldots ,1)-\frac{1}{KL}\max _{j=1}^{N}\Vert v_{\mathrm {int},j}\Vert _{L^{1}({\mathbb {R}})}\succeq -(\varepsilon ,\ldots ,\varepsilon ). \end{aligned}$$

A similar argument shows that

$$\begin{aligned} \frac{1}{KL}\int _{0}^{KL}{\mathbf {v}}_{+}(x)\,{d}x-{\mathbf {a}}\preceq (\varepsilon ,\ldots ,\varepsilon ). \end{aligned}$$

Since this is possible for every \(\varepsilon >0\), we have \({\mathbf {v}}\in {\mathscr {B}}_{{\mathbf {a}}}\).    \(\square \)

Weighted Spaces

We now record several useful results on weighted spaces. We begin with some basic lemmas that are used throughout the paper. Then we will prove bounds on the heat kernel on weighted spaces.

Basic properties of weighted spaces

Our first lemma allows us to upgrade convergence from one weight to another.

Proposition B.1

Let \({\mathcal {T}}\) be a metric space and \(w_{1},w_{2},w_{3}\) be weights on \({\mathbb {R}}\) so that

$$\begin{aligned} \lim \limits _{|x|\rightarrow \infty }\frac{w_{1}(x)}{w_{2}(x)}=0. \end{aligned}$$

Suppose that \(u_n\in {\mathcal {C}}_{\mathrm {b}}({\mathcal {T}};{\mathcal {C}}_{w_{1}}({\mathbb {R}}))\), and \(u\in {\mathcal {C}}_{\mathrm {b}}({\mathcal {T}};{\mathcal {C}}_{w_{3}}({\mathbb {R}}))\) satisfy

$$\begin{aligned} \lim \limits _{n\rightarrow \infty } \Vert u_{n}-u\Vert _{{\mathcal {C}}_{\mathrm {b}}({\mathcal {T}};{\mathcal {C}}_{w_{3}} ({\mathbb {R}}))}=0 \quad \text {and} \quad \sup \limits _{n\in {\mathbb {N}}}\Vert u_{n}\Vert _{{\mathcal {C}}_{\mathrm {b}}({\mathcal {T}} ;{\mathcal {C}}_{w_{1}}({\mathbb {R}}))}<\infty . \end{aligned}$$

Then \(u\in {\mathcal {C}}_{\mathrm {b}}({\mathcal {T}};{\mathcal {C}}_{w_{2}}({\mathbb {R}}))\), and \(\lim \limits _{n\rightarrow \infty }\Vert u_{n}-u\Vert _{{\mathcal {C}}_{\mathrm {b}}({\mathcal {T}};{\mathcal {C}}_{w_{2}}({\mathbb {R}}))}=0\) as well.


Fix \(\varepsilon >0\) and define

$$\begin{aligned} K = \sup \limits _{n\in {\mathbb {N}}} \Vert u_{n}\Vert _{{\mathcal {C}}_{\mathrm {b}}({\mathcal {T}};{\mathcal {C}}_{w_{1}}({\mathbb {R}}))}<\infty . \end{aligned}$$

Then choose M so that

$$\begin{aligned} \left| \frac{w_{1}(x)}{w_{2}(x)}\right| \le \frac{\varepsilon }{2K}~~~\hbox { if }|x|\ge M, \end{aligned}$$

and N so large that if \(n\ge N\) then

$$\begin{aligned} \Vert u_{n}-u\Vert _{{\mathcal {C}}_{\mathrm {b}}({\mathcal {T}};{\mathcal {C}}_{w_{3}}({\mathbb {R}}))} \le \varepsilon \inf \limits _{|x|\le M}\frac{w_{2}(x)}{w_{3}(x)}. \end{aligned}$$

Now, for any \(n\ge N\), if \(|y|\le M\), then for all \(t\in {\mathcal {T}}\) we have

$$\begin{aligned} |(u_{n}-u)(t,y)|\le \varepsilon w_{3}(y)\inf \limits _{x\in [-M,M]}\frac{w_{2}(x)}{w_{3}(x)}\le \varepsilon w_{2}(y) \end{aligned}$$

while if \(|y|\ge M\) we have

$$\begin{aligned} |(u_{n}-u)(t,y)|\le \varepsilon w_{3}(y)\inf \limits _{x\in [-M,M]}\frac{w_{2}(x)}{w_{3}(x)}\le \varepsilon w_{2}(y). \end{aligned}$$

Therefore, \(\Vert u_{n}-u\Vert _{{\mathcal {C}}_{\mathrm {b}}({\mathcal {T}};{\mathcal {C}}_{w_{2}} ({\mathbb {R}}))}<\varepsilon \). This proves that \(\lim \limits _{n\rightarrow \infty }\Vert u_{n}-u\Vert _{{\mathcal {C}}_{\mathrm {b}}({\mathcal {T}}; {\mathcal {C}}_{w_{2}}({\mathbb {R}}))}=0\), as claimed. \(\quad \square \)

We next establish a form of the Arzelà–Ascoli theorem in weighted spaces.

Proposition B.2

Suppose that \(w_{1},w_{2},w_{3}\) are weights so that \(\lim \limits _{|x|\rightarrow \infty }\frac{w_{1}(x)}{w_{2}(x)}=0\) and fix \(\alpha > 0\). Then the embedding

$$\begin{aligned} {\mathcal {C}}_{w_{1}}({\mathbb {R}})\cap {\mathcal {C}}_{w_{3}}^{\alpha } ({\mathbb {R}})\hookrightarrow {\mathcal {C}}_{w_{2}}({\mathbb {R}}) \end{aligned}$$

is compact, where \({\mathcal {C}}_{w_{1}}({\mathbb {R}})\cap {\mathcal {C}}_{w_{3}}^{\alpha }({\mathbb {R}})\) is equipped with the norm \(\Vert u\Vert _{{\mathcal {C}}_{w_{1}}({\mathbb {R}})\cap {\mathcal {C}}_{w_{3}}^{\alpha } ({\mathbb {R}})}=\Vert u\Vert _{{\mathcal {C}}_{w_{1}}({\mathbb {R}})} +\Vert u\Vert _{{\mathcal {C}}_{\mathrm {p}_{w_{3}}}^{\alpha }({\mathbb {R}})}\).


It suffices to show that the unit ball of \({\mathcal {C}}_{w_{1}}({\mathbb {R}})\cap {\mathcal {C}}_{w_{3}}^{\alpha }({\mathbb {R}})\) is precompact in \({\mathcal {C}}_{w_{2}}({\mathbb {R}})\). Fix a sequence \((v_{n})_{n}\) of elements of this unit ball. On any compact subset of \({\mathbb {R}}\), \((v_{n})\) is uniformly bounded and Hölder. Thus by Arzelà–Ascoli and diagonalization, there exists a subsequence \((v_{n_{k}})_{k}\) which converges locally uniformly to some \(v\in {\mathcal {C}}({\mathbb {R}})\). As noted in the proof of Lemma B.4, this is equivalent to convergence in some weighted space. Since \((v_{n})\) is uniformly bounded in \({\mathcal {C}}_{w_{1}}({\mathbb {R}})\), Proposition B.1 implies that \(v\in {\mathcal {C}}_{w_{2}}({\mathbb {R}})\) and \(\lim \limits _{k\rightarrow \infty }v_{n_{k}}=v\) in \({\mathcal {C}}_{w_{2}}({\mathbb {R}})\). \(\quad \square \)

Finally, we record a compactness criterion in \({\mathcal {X}}_m\).

Lemma B.3

If \(K\subset {\mathcal {X}}_{m}\) is such that K is compact in the topology of \({\mathcal {C}}_{\mathrm {p}_{\ell }}({\mathbb {R}})\) for each \(\ell >m\), then K is compact in the topology of \({\mathcal {X}}_{m}\) as well.


Let \((v_{n})_{n}\) be a sequence of elements in K. By a diagonal argument, there is a subsequence \((v_{n_k})_k\) which converges in the topology of \({\mathcal {C}}_{\mathrm {p}_{\ell }}({\mathbb {R}})\) for all \(\ell >m\), and hence in the topology of \({\mathcal {X}}_{m}\). \(\quad \square \)

Heat kernel bounds in weighted spaces

Here, we prove some weighted estimates for the heat kernel.

Lemma B.4

Fix a weight \(w\in \{(\log \langle \cdot \rangle )^{3/4}\}\cup \{\mathrm {p}_{\ell }{\ :\ }\ell \in {\mathbb {R}}\}\), \(\beta \ge \alpha \ge 0\), and \(T<\infty \). There is a constant \(C=C(w,\alpha ,\beta ,T)<\infty \) so that for all \(t\in (0,T]\) and \(f\in {\mathcal {C}}_{w}^{\alpha }({\mathbb {R}})\) we have

$$\begin{aligned} \Vert G_{t}*f\Vert _{{\mathcal {C}}_{w}^{\beta }({\mathbb {R}})}\le Ct^{-\frac{\beta -\alpha }{2}}\Vert f\Vert _{{\mathcal {C}}_{w}^{\alpha }({\mathbb {R}})} \end{aligned}$$

In particular,

$$\begin{aligned} \Vert \partial _{x}G_{t}*f\Vert _{{\mathcal {C}}_{w}^{\beta }({\mathbb {R}})}\le Ct^{-\frac{\beta -\alpha +1}{2}}\Vert f\Vert _{{\mathcal {C}}_{w}^{\alpha }({\mathbb {R}})}. \end{aligned}$$

In the case \(\alpha =0\), it is only necessary to assume that \(f\in L_{w}^{\infty }({\mathbb {R}})\) and the norm \(\Vert f\Vert _{{\mathcal {C}}_{w}^{\alpha }({\mathbb {R}})}\) can be replaced by \(\Vert f\Vert _{L_{w}^{\infty }({\mathbb {R}})}\) on the right-hand sides of (B.1) and (B.2).

The proof of this lemma is word-for-word the same as that of [36, Lemma 2.8]. There, only exponential weights (since a uniformity statement in the weight is needed) and continuous functions are considered, but there is no difference in the treatment given the Gaussian decay of the heat kernel. The essence of the argument is that only singularity in the heat kernel is at \(t=0\), \(x=0\), so the part of the heat kernel that is exposed to the growth of f at infinity is smooth, and moreover decays quickly enough not to pose any difficulty for these estimates.

Lemma B.5

Fix \(m \in {\mathbb {R}}\). If \(p \in [1, \infty )\) and \(f \in L_{\mathrm {p}_m}^p({\mathbb {R}})\), then \(G_t *f \rightarrow f\) in \(L_{\mathrm {p}_m}^p({\mathbb {R}})\) as \(t \downarrow 0\). If \(f \in L_{\mathrm {p}_m}^\infty ({\mathbb {R}})\), then \(G_t *f \overset{\mathrm {w}^*}{\longrightarrow } f\) in \(L_{\mathrm {p}_m}^\infty ({\mathbb {R}})\) as \(t \downarrow 0\). Finally, if \(f \in {\mathcal {C}}_{\mathrm {p}_{m}}({\mathbb {R}})\), then \(G_t *f \rightarrow f\) in \({\mathcal {X}}_m\) as \(t \downarrow 0\).


First fix \(p \in [1, \infty )\) and \(f \in L_{\mathrm {p}_m}^p({\mathbb {R}})\). We provide a simple variant of a standard “approximation of the identity” argument to deal with the weighted spaces. Using the scaling symmetry of G, we can write

$$\begin{aligned} G_{t}*f(x)-f(x)=\int _{{\mathbb {R}}}[f(x-\sqrt{t}y)-f(x)]G_{1}(y)\,{d}y, \end{aligned}$$

so by the triangle inequality,

$$\begin{aligned} \Vert G_{t}*f-f\Vert _{L_{\mathrm {p}_m}^{p}({\mathbb {R}})}\le \int _{{\mathbb {R}}}\Vert \tau _{\sqrt{t}y}f-f\Vert _{L_{\mathrm {p}_m}^{p}({\mathbb {R}})}G_{1}(y)\,{d}y. \end{aligned}$$

We will use the dominated convergence theorem, so we first establish an integrable majorant. Assume \(t\in (0,1]\). We can easily verify that there exists \(C = C(m) <\infty \) such that

$$\begin{aligned} \mathrm {p}_{-m}(a + b) \le C \mathrm {p}_{-m}(a) \mathrm {p}_{|m|}(b) \end{aligned}$$

for all \(a, b \in {\mathbb {R}}\). In the sequel, we permit C to change from line to line. Then

$$\begin{aligned} \Vert \tau _{\sqrt{t}y}f\Vert _{L_{\mathrm {p}_m}^{p}({\mathbb {R}})} = \left( \int _{{\mathbb {R}}} |f(x)|^p \mathrm {p}_{m}(x+\sqrt{t}y)^{-p}\,{d}x\right) ^{\frac{1}{p}} \le C\mathrm {p}_{|m|}(y)\Vert f\Vert _{L_{\mathrm {p}_{m}}^{p}({\mathbb {R}})}. \end{aligned}$$

Since \(\mathrm {p}_{|m|}\) is integrable against the Gaussian \(G_{1}\), this is a suitable majorant.

By the dominated convergence theorem, it now suffices to prove pointwise (in y) convergence to 0 as \(t\downarrow 0\) in (B.3). We can therefore fix \(y \in {\mathbb {R}}\) and consider \(t > 0\) such that \(\sqrt{t} y \le 1\). For fixed \(\varepsilon >0\), we can find a compactly-supported continuous function \(\zeta \) on \({\mathbb {R}}\) so that \(\Vert \zeta -f\Vert _{L_{\mathrm {p}_{m}}^{p}({\mathbb {R}})} < \varepsilon \). Then we have

$$\begin{aligned} \Vert \tau _{\sqrt{t}y}f-f\Vert _{L_{\mathrm {p}_{m}}^{p}({\mathbb {R}})}\le \Vert \tau _{\sqrt{t}y}f-\tau _{\sqrt{t}y}\zeta \Vert _{L_{\mathrm {p}_{m}}^{p}({\mathbb {R}})} +\Vert \tau _{\sqrt{t}y}\zeta -\zeta \Vert _{L_{\mathrm {p}_{m}}^{p}({\mathbb {R}})} +\Vert \zeta -f\Vert _{L_{\mathrm {p}_{m}}^{p}({\mathbb {R}})}. \end{aligned}$$

By (B.4) and \(\sqrt{t} y \le 1\), the first and third terms are each less than a constant times \(\varepsilon \) and the second term goes to 0 as \(t \rightarrow 0\). Therefore \(G_t *f \rightarrow f\) in \(L_{\mathrm {p}_m}^p({\mathbb {R}})\) as \(t \downarrow 0\).

Now suppose \(f \in L_{\mathrm {p}_m}^\infty ({\mathbb {R}})\), and fix \(\phi \) in the dual space \(L_{\mathrm {p}_{-m}}^{1}({\mathbb {R}})\). We must show that

$$\begin{aligned} \langle \phi ,G_{t}*f\rangle \rightarrow \langle \phi ,f\rangle \quad \text {as } t \downarrow 0. \end{aligned}$$

Since \(G_{t}\) is symmetric, we have \(\langle \phi ,G_{t}*f\rangle =\langle G_{t}*\phi ,f\rangle \). But we have just shown that \(G_{t}*\phi \rightarrow \phi \) in \(L_{\mathrm {p}_{-m}}^1({\mathbb {R}})\), so (B.5) follows.

Finally, suppose that \(f \in {\mathcal {C}}_{\mathrm {p}_{m}}({\mathbb {R}})\). Fix \(\varepsilon > 0\) and \(x \in {\mathbb {R}}\). We write

$$\begin{aligned} G_t *f(x) - f(x) = \int _{{\mathbb {R}}}[f(y)-f(x)]G_t(x-y)\,{d}y. \end{aligned}$$

Now \(|f(y)|\le \Vert f\Vert _{{\mathcal {C}}_{\mathrm {p}_{m}}({\mathbb {R}})}\mathrm {p}_{m}(y)\) and \(\mathrm {p}_{m} G_t(x - \cdot )\in L^{1}({\mathbb {R}})\). When \(t \in (0, 1]\), \(G_t\) is decreasing in t outside a compact set. Thus there exists a compact set \(K\subset {\mathbb {R}}\) containing x so that

$$\begin{aligned} \int _{{\mathbb {R}}\setminus K}|f(y)-f(x)|G_t(x-y)\,{d}y<\varepsilon \end{aligned}$$

for all \(n\ge 0\). On K, f is uniformly continuous and bounded. Thus there exists a \(\delta >0\) such that \(|f(y)-f(x)|<\varepsilon \) when \(|y-x|<\delta \). Since \(G_t\) has unit mass, this implies

$$\begin{aligned} \int _{B_{\delta }(x)}|f(y)-f(x)|G_t(x-y)\,{d}y<\varepsilon \end{aligned}$$

for all \(t \in (0, 1]\). Finally, \(G_t(x-y)\rightarrow 0\) uniformly on \(K\setminus B_{\delta }(x)\) as \(t \downarrow 0\), so there exists \(\delta > 0\) such that

$$\begin{aligned} \int _{K\setminus B_{\delta }(x)}|f(y)-f(x)|G_t(x-y)\,{d}y<\varepsilon \end{aligned}$$

for all \(t < \delta \). Together, these bounds show that \(|G_t *f(x) - f(x)|\rightarrow 0\) as \(n\rightarrow 0\).

In fact, the convergence is locally uniform in x. But locally uniform convergence is equivalent to the existence of a weight w such that

$$\begin{aligned} \lim _{n\rightarrow \infty }\Vert G_t *f - f\Vert _{{\mathcal {C}}_{w}({\mathbb {R}})}=0. \end{aligned}$$

Combining this with the uniform bound (B.1) with \(\alpha = \beta = 0\) and \(w = \mathrm {p}_m\), Proposition B.1 implies

$$\begin{aligned} \lim _{n\rightarrow \infty }\Vert G_t *f - f\Vert _{{\mathcal {C}}_{\mathrm {p}_{\ell }}({\mathbb {R}})}=0 \end{aligned}$$

for any \(\ell >m\). That is, \(G_t *f \rightarrow f\) in \({\mathcal {X}}_m\). \(\quad \square \)

Next, we show an estimate with super-exponential weights. The restriction \(\beta <2\) is needed in the following lemma simply because the heat equation is not well-posed for initial conditions growing like \(\exp (cx^{2})\) with \(c>0\).

Lemma B.6

Fix \(\beta \in [3/2,2)\) and define, for \(\lambda \ge 0\), \(\mathrm {q}_{\lambda }(x)=\mathrm {e}^{\lambda \langle x\rangle ^{\beta }}\). For any \(\Lambda >0\) and \(T>0\), there exists a \(C<\infty \) so that for all \(\lambda \in [0,\Lambda ]\), \(t\in (0,T]\), \(f\in {\mathcal {C}}_{q_{\lambda }}({\mathbb {R}})\), and \(x\in {\mathbb {R}}\), we have

$$\begin{aligned} |\partial _{x}G_{t}*f(x)|\le Ct^{-\frac{1}{2}}\mathrm {e}^{Ct\langle x\rangle ^{2(\beta -1)}}q_{\lambda }(x)\Vert f\Vert _{{\mathcal {C}}_{\mathrm {q}_{\lambda }} ({\mathbb {R}})}. \end{aligned}$$

Remark B.7

This also holds for \(\beta \in (1,3/2)\). The argument is similar but not identical, and is not needed in this paper, so we omit it.


Throughout the proof, C denotes a positive constant that depends only on \(\Lambda \) and T. It may change from line to line. We may assume without loss of generality that \(\Vert f\Vert _{{\mathcal {C}}_{\mathrm {q}_{\lambda }}({\mathbb {R}})}=1\). We begin by noting that \(|y|\le \exp (y^{2}/4)\) for all \(y\in {\mathbb {R}}\), so

$$\begin{aligned} |\partial _{x}G_{t}|\le \sqrt{2}t^{-\frac{1}{2}}G_{2t}. \end{aligned}$$

Therefore, we have

$$\begin{aligned} |\partial _{x}G_{t}*f(x)|\le \sqrt{2}t^{-\frac{1}{2}}G_{2t}*\mathrm {q}_{\lambda }(x), \end{aligned}$$

so it remains to bound \(|G_{2t}*\mathrm {q}_{\lambda }(x)|\). The function \(w(s,\cdot ) = G_{s}*q_{\lambda }\) solves the heat equation with initial condition \(\mathrm {q}_{\lambda }\), so we can bound it from above by constructing a supersolution v with the same initial condition. Set

$$\begin{aligned} v(s,x)=\exp \left( As\langle x\rangle ^{2(\beta -1)}+Bs^{\frac{\beta }{2-\beta }}\right) q_{\lambda }(x) \end{aligned}$$

for constants \(A,B>0\) to be determined. Then we have

$$\begin{aligned} \partial _{s}v\ge \left( A\langle x\rangle ^{2(\beta -1)}+Bs^{\frac{2(\beta -1)}{2-\beta }}\right) v \end{aligned}$$


$$\begin{aligned} \partial _{xx}v\le 8\left( \lambda ^{2}+\lambda +A(\lambda +1)s\langle x \rangle ^{\beta -2}+A^{2}s^{2}\langle x\rangle ^{2(\beta -2)}\right) \langle x\rangle ^{2(\beta -1)}v. \end{aligned}$$

Comparing these, in order for v to be a supersolution for the heat equation, i.e. to satisfy

$$\begin{aligned} \partial _{s}v\ge \frac{1}{2}\partial _{xx}v, \end{aligned}$$

it suffices to choose A and B so that

$$\begin{aligned} 4\left( \lambda ^{2}+\lambda +A(\lambda +1)s\langle x\rangle ^{\beta -2}+A^{2}s^{2}\langle x\rangle ^{2(\beta -2)}\right) \le A+Bs^{\frac{2(\beta -1)}{2-\beta }}\langle x\rangle ^{-2(\beta -1)}.\nonumber \\ \end{aligned}$$

To accomplish this, let \(A=4(\lambda ^{2}+\lambda )+2\) and \(\xi =s\langle x\rangle ^{\beta -2}\). Then for (B.10) to hold, it suffices to choose B so that

$$\begin{aligned} 4(\lambda +1)A\xi +4A^{2}\xi ^{2}\le 2+B\xi ^{\frac{2(\beta -1)}{2-\beta }} \end{aligned}$$

for all \(\xi \ge 0\). When \(\xi \le 1/(4A(\lambda +1))\), the left side of (B.11) is bounded by 2, and the inequality holds regardless of B. Moreover, \(2(\beta -1)/(2-\beta )\ge 2\) since \(\beta \ge 3/2\), so the right side of (B.11) grows at least as fast as the left as \(\xi \rightarrow +\infty \). Thus, there exists \(B=B(\Lambda )\) sufficiently large that (B.11) holds also when \(\xi \ge 1/(4A(\lambda +1))\). With these values of A and B, v satisfies (B.9) and \(v(0,\cdot )\equiv \mathrm {q}_{\lambda }\equiv w(0,\cdot )\). Thus \(w(s,x)\le v(s,x)\) for all \(s\ge 0\) and \(x\in {\mathbb {R}}\) by the comparison principle for the heat equation. The bound (B.6) then follows from (B.7) and (B.8). \(\quad \square \)

Classical Solutions are Mild

In this appendix we prove a converse to Lemma 2.8.

Lemma C.1

(Classical solutions are mild). Suppose that \(m\in (0,1)\), \(L\in (0,\infty ]\), \(T>0\) and \(\theta ^{[L]}\in \tilde{{\mathcal {Z}}}_{m,T}\) is a classical solution to (2.14). Then \(\theta ^{[L]}\) satisfies (2.21).


Let \(f=-\partial _{x}(\theta ^{[L]}+\psi ^{[L]})^{2}\) denote the nonlinear forcing in (2.14). Fix \(t\in (0,T]\) and define, for \(s\in [0,t]\),

$$\begin{aligned} \Theta (s,x)&=G_{s}*\theta ^{[L]}(t-s,x),&F(s,x)&=G_{s}*f(t-s,x). \end{aligned}$$

It is clear that \(\Theta \) is twice-differentiable in space. We claim that it is also continuous in s, pointwise in x. Fix \(\ell >m\), \((s,x)\in [0,t]\times {\mathbb {R}}\), and a sequence \((s_{n})_{n \in {\mathbb {N}}}\subset (0,t)\) converging to s. If \(s > 0\), we use

$$\begin{aligned} \begin{aligned} |\Theta (s_{n},x)-\Theta (s,x)|&\le |(G_{s_{n}}-G_{s})*\theta ^{[L]}(t-s_{n},\cdot )(x)|\\&\quad +|G_{s}*[\theta ^{[L]}(t-s_{n},\cdot )-\theta ^{[L]}(t-s,\cdot )](x)|. \end{aligned} \end{aligned}$$

Then \(\Vert G_{s_{n}}-G_{s}\Vert _{L_{\mathrm {p}_{-\ell }}^{1}({\mathbb {R}})}\rightarrow 0\). Since \(\theta ^{[L]}\) is uniformly bounded in \(L_{\mathrm {p}_{\ell }}^{\infty }({\mathbb {R}})\), we have

$$\begin{aligned} |(G_{s_{n}}-G_{S})*\theta ^{[L]}(t-s_{n},\cdot )(x)|\le \Vert G_{s_{n}}-G_{s} \Vert _{L_{\mathrm {p}_{-\ell }}^{1}({\mathbb {R}})}\Vert \theta ^{[L]}(t-s_{n},\cdot ) \Vert _{L_{\mathrm {p}_{\ell }}^{\infty }({\mathbb {R}})}\rightarrow 0. \end{aligned}$$

On the other hand,

$$\begin{aligned}&G_{s}*[\theta ^{[L]}(t-s_{n},\cdot )-\theta ^{[L]}(t-s,\cdot )](x)\\&\quad =\int _{{\mathbb {R}}}G_{s}(x-y)[\theta ^{[L]}(t-s_{n},y)-\theta ^{[L]}(t-s,y)]\,{d}y\rightarrow 0 \end{aligned}$$

by the weak-\(*\) continuity of \(\theta ^{[L]}\), since \(G_{s}\in L_{\mathrm {p}_{-\ell }}^{1}({\mathbb {R}})\). By (C.1), \(|\Theta (s_{n},x)-\Theta (s,x)|\rightarrow 0\) as \(n\rightarrow \infty \).

Now suppose \(s=0\). Then \(G_{0}=\delta _{0}\) is singular, so we must argue differently. In this case, we are considering \(\theta ^{[L]}\) near a time \(t>0\), where it is continuous. Fix \(\varepsilon >0\), and consider the opposite decomposition

$$\begin{aligned} \begin{aligned} |\Theta (s_{n},x)-\Theta (s,x)|&\le |G_{s_{n}}*[\theta ^{[L]}(t-s_{n},\cdot )- \theta ^{[L]}(t,\cdot )](x)|\\&\qquad +|(G_{s_{n}}-\delta _{0})*\theta ^{[L]}(t,\cdot )(x)|. \end{aligned} \end{aligned}$$

Since \(\theta ^{[L]}\in {\mathcal {C}}_{\mathrm {b}}((0,S];{\mathcal {C}}_{\mathrm {p}_{\ell }} ({\mathbb {R}}))\), there exists a \(\delta >0\) such that

$$\begin{aligned} \Vert \theta ^{[L]}(t-s_{n},\cdot )-\theta ^{[L]}(t,\cdot )\Vert _{{\mathcal {C}}_{\mathrm {p}_{\ell }} ({\mathbb {R}})}<\varepsilon \end{aligned}$$

when \(s_{n}<\delta \). Now \(G_{s_{n}}*\mathrm {p}_{\ell }\le C_\ell \mathrm {p}_{\ell }\), so

$$\begin{aligned} |G_{s_{n}}*[\theta ^{[L]}(t-s_{n},\cdot )-\theta ^{[L]}(t,\cdot )](x)|\le C_\ell \varepsilon \mathrm {p}_{\ell }(x). \end{aligned}$$

For the second term of (C.2), we use the fact that \(\theta ^{[L]}(t,\cdot )\in {\mathcal {C}}_{\mathrm {p}_{\ell }}({\mathbb {R}})\), so by Lemma B.5, we have pointwise convergence and \(|(G_{s_{n}}-\delta _{0})*\theta ^{[L]}(t,\cdot )(x)|\rightarrow 0\). Thus by (C.2), \(|\Theta (s_{n},x)-\Theta (0,x)|\rightarrow 0\) as \(n\rightarrow \infty \), as desired.

Next, (2.9) implies that \(\Theta \) is differentiable in s on (0, t) and that

$$\begin{aligned} \partial _s \Theta (s, x)&= (\partial _s G_s) *\theta ^{[L]}(t - s, x) - G_s *\partial _t \theta ^{[L]}(t - s, x)\\&= \frac{1}{2} \partial _{xx} G_s *\theta ^{[L]}(t - s, x) - \frac{1}{2} G_s *\partial _{xx}\theta ^{[L]}(t - s, x) - F(s, x). \end{aligned}$$

Now \(\theta ^{[L]}(t - s, \cdot )\) is a tempered distribution, so we may exchange differentiation and convolution to find

$$\begin{aligned} \partial _s \Theta (s, x) = -F. \end{aligned}$$

The continuity of \(\Theta \) in time ensures that

$$\begin{aligned} \Theta (t,x)-\Theta (0,x)=\int _{0}^{t}\partial _{s}\Theta (s,x)\,{d}s = - \int _0^t F(s, x) \,{d}s \end{aligned}$$

for all \(x\in {\mathbb {R}}\). After a change of variables in the integral, this is simply (2.21). \(\quad \square \)

Elementary Probabilistic and Analytic Lemmas

In this appendix we prove some elementary technical lemmas that were deferred until this point to avoid disrupting the flow of the paper.

Several symmetry arguments in the paper relied on the following lemma.

Lemma D.1

Let \(X_{1}\) and \(X_{2}\) be random variables such that \(X_{1}\overset{\mathrm {law}}{=}X_{2}\) and \({\mathbb {E}}(X_{2}-X_{1})^{-}>-\infty \). Then \({\mathbb {E}}|X_{2}-X_{1}|<\infty \) and \({\mathbb {E}}(X_{2}-X_{1})=0\).


Of course the claim is obvious if \({\mathbb {E}}|X_{i}|<\infty \), but for our applications in the paper it will be convenient to not have to assume this. Define \(f_{M}(x)=\max \{\min \{x,M\},-M\}\) and note that \(f_{M}\) is bounded and 1-Lipschitz. The function \(g_{M}(x)=f_{M}(x)-x\) is also 1-Lipschitz. Fix \(R\in (0,\infty ]\) and compute

$$\begin{aligned}&{\mathbb {E}}(|g_{M}(X_{2})-g_{M}(X_{1})| \cdot {\mathbf {1}}\{X_{2}-X_{1}\le R\})\nonumber \\&\quad ={\mathbb {E}}(|g_{M}(X_{2})-g_{M}(X_{1})|{\mathbf {1}}\{\max \{|X_{1}|,|X_{2}|\}\ge M\}{\mathbf {1}}\{X_{2}-X_{1}\le R\})\nonumber \\&\quad \le {\mathbb {E}}(|X_{2}-X_{1}|{\mathbf {1}}\{\max \{|X_{1}|,|X_{2}|\}\ge M\}{\mathbf {1}}\{X_{2}-X_{1}\le R\}). \end{aligned}$$

For each \(0<R<\infty \) fixed, the last expression in (D.1) goes to 0 as \(M\rightarrow \infty \) by the dominated convergence theorem, since \({\mathbb {E}}(|X_{2}-X_{1}|{\mathbf {1}}\{X_{2}-X_{1}\le R\})\) is finite since \((X_{2}-X_{1})^{+}{\mathbf {1}}\{X_{2}-X_{1}\le R\}\) is bounded and \({\mathbb {E}}(X_{2}-X_{1})^{-}\) is finite by assumption. Therefore, we have

$$\begin{aligned} 0&=\lim _{M\rightarrow \infty }{\mathbb {E}}(|g_{M}(X_{2})-g_{M}(X_{1})| \cdot {\mathbf {1}}\{X_{2}-X_{1}\le R\})\nonumber \\&=\lim _{M\rightarrow \infty }{\mathbb {E}}(|f_{M}(X_{2})-f_{M}(X_{1}) -(X_{2}-X_{1})|\cdot {\mathbf {1}}\{X_{2}-X_{1}\le R\})\nonumber \\&\ge \limsup _{M\rightarrow \infty }\big |{\mathbb {E}}((f_{M}(X_{2}) -f_{M}(X_{1})){\mathbf {1}}\{X_{2}-X_{1}\le R\})\nonumber \\&\qquad -{\mathbb {E}}((X_{2}-X_{1}){\mathbf {1}}\{X_{2}-X_{1}\le R\})\big |. \end{aligned}$$

Since \(X_{1}\overset{\mathrm {law}}{=}X_{2}\), we have \({\mathbb {E}}[f_{M}(X_{2})-f_{M}(X_{1})]=0\). Also, \(f_{M}(x)\) is monotone in x, so

$$\begin{aligned}&{\mathbb {E}}[(f_{M}(X_{2})-f_{M}(X_{1})){\mathbf {1}}\{X_{2}-X_{1}\le R\}]\\&\quad =-{\mathbb {E}}[(f_{M}(X_{2})-f_{M}(X_{1})){\mathbf {1}}\{X_{2}-X_{1}>R\}]\le 0 \end{aligned}$$

for all \(M,R\in (0,\infty )\). Using this in (D.2) gives

$$\begin{aligned} {\mathbb {E}}((X_{2}-X_{1}){\mathbf {1}}\{X_{2}-X_{1}\le R\})\le 0 \end{aligned}$$

for all \(R\in (0,\infty )\). Taking \(R\rightarrow +\infty \) and using the monotone convergence theorem and the assumption \({\mathbb {E}}(X_{2}-X_{1})^{-}>-\infty \) yields \({\mathbb {E}}(X_{1}-X_{2})\le 0\), so in particular \({\mathbb {E}}|X_{1}-X_{2}|<\infty \). Thus we can take \(R=+\infty \) in (D.1) and again use the dominated convergence theorem to obtain (D.2) with \(R=+\infty \), namely

$$\begin{aligned} 0\ge \limsup _{M\rightarrow \infty }\left| {\mathbb {E}}[f_{M}(X_{2})-f_{M}(X_{1})] -{\mathbb {E}}(X_{2}-X_{1})\right| =|{\mathbb {E}}(X_{2}-X_{1})|, \end{aligned}$$

and so \({\mathbb {E}}(X_{2}-X_{1})=0\) as claimed. \(\quad \square \)

The following lemma will be used in the proof of Lemma D.3 below.

Lemma D.2

Let \(\eta \in {\mathcal {C}}^{3}(I)\) for some closed interval \(I \subset {\mathbb {R}}\) and define \(A=\Vert \eta \Vert _{{\mathcal {C}}^{3}(I)}\). If \(x_{1}\ne x_{2}\in I\) satisfy \(\eta (x_{1})=\eta (x_{2})\) and \(\eta '(x_{1}),\eta '(x_{2})>\varepsilon \) or \(\eta '(x_{1}),\eta '(x_{2})<-\varepsilon \), then \(|x_{1}-x_{2}|\ge \sqrt{2\varepsilon /A}.\)


Without loss of generality, we may assume that \(x_2 > x_1\) and \(\eta '(x_{1}),\eta '(x_{2})>\varepsilon \). Let \(\delta =x_{2}-x_{1}\). By Rolle’s theorem, there is a \(y\in (x_{1},x_{2})\) so that \(\eta '(y)=0\). By the mean value theorem, there exist \(z_{1}\in (x_{1},y)\) and \(z_{2}\in (y,x_{2})\) so that

$$\begin{aligned} \eta ''(z_{1})&\le -\delta ^{-1}\varepsilon ,&\eta ''(z_{2})&\ge \delta ^{-1}\varepsilon . \end{aligned}$$

By another application of the mean value theorem, there exists a \(w\in (z_{1},z_{2})\) so that

$$\begin{aligned} \eta '''(w)\ge 2\delta ^{-2}\varepsilon . \end{aligned}$$

This means that \(2\delta ^{-2}\varepsilon \le A\), so \(\delta \ge \sqrt{2\varepsilon /A}\), as claimed. \(\quad \square \)

The following lemma was used in the proof of Lemma 3.5.

Lemma D.3

Suppose that \(\zeta \ge 0\) is in the Schwartz class and \(\eta \) is a smooth function all of whose derivatives have at most polynomial growth at infinity. Define

$$\begin{aligned} g(\lambda )=\sum _{\begin{array}{c} y\in \eta ^{-1}(\lambda )\\ \eta '(y)\ne 0 \end{array} }|\eta '(y)|\zeta (y). \end{aligned}$$

Then g is a continuous function of \(\lambda \).


Let \(\zeta _{k} \ge 0\) be smooth such that \({\text {supp}}\zeta _{k}\subset [k-1,k+2]\) and \(\sum \limits _{k\in {\mathbb {Z}}}\zeta _{k}=\zeta \). Let

$$\begin{aligned} g_{k}(\lambda )=\sum _{\begin{array}{c} y\in \eta ^{-1}(\lambda )\\ \eta '(y)\ne 0 \end{array} }|\eta '(y)|\zeta _{k}(y). \end{aligned}$$

We first show that \(g_{k}\) is continuous. Let \(A_{k}=\Vert \eta \Vert _{{\mathcal {C}}^{3}([k-1,k+2])}+\Vert \zeta _{k}\Vert _{{\mathcal {C}}^{0}({\mathbb {R}})} +\Vert \eta '\zeta _k\Vert _{{\mathcal {C}}^1({\mathbb {R}})}\) + 1. Define

$$\begin{aligned} S_{k,\ell }(\lambda )=\{y\in (k-1,k+2) \mid \eta (y)=\lambda ,|\eta '(y)|\in [2^{-\ell },2^{-\ell +1})\}. \end{aligned}$$

Lemma D.2 implies

$$\begin{aligned} |S_{k,\ell }(\lambda )|\le 2^3 A_{k}^{1/2}2^{\ell /2}, \end{aligned}$$


$$\begin{aligned} \sum _{y\in S_{k,\ell }(\lambda )}|\eta '(y)|\zeta _{k}(y)\le 2^{-\ell + 1} \Vert \zeta _k\Vert _{{\mathcal {C}}^0({\mathbb {R}})} |S_{k, \ell }(\lambda )| \le 2^4 A_{k}^{3/2} 2^{-\ell /2} \end{aligned}$$

for all \(\lambda \).

Now fix \(\varepsilon >0\) and choose \(\ell \) so large that \(2^9 A_k^{3/2} 2^{-\ell /2} < \varepsilon .\) Define

$$\begin{aligned} T_{k, \ell }^{+}(\lambda )&=\{y\in (k - 1, k + 2) \mid \eta (y)=\lambda , \eta '(y)\ge 2^{-\ell }\}, \\ T_{k, \ell }^{-}(\lambda )&=\{y\in (k - 1, k + 2) \mid \eta (y)=\lambda , \eta '(y)\le -2^{-\ell }\}. \end{aligned}$$

Suppose that \(\lambda _{1}<\lambda _{2}\) satisfy \(\lambda _{2}-\lambda _{1} < 2^{-2\ell - 2}A_k^{-1}\). Take \(x\in T_{k, \ell }^{+}(\lambda _{1})\). On the interval \({[x - 2^{-\ell -1} A_k^{-1}, x + 2^{-\ell -1} A_k^{-1}]}\), we must have

$$\begin{aligned} \eta ' \ge \eta '(x) - \Vert \eta ''\Vert _{{\mathcal {C}}^0([k-1, k+2])} 2^{-\ell -1} A_k^{-1} \ge 2^{-\ell - 1}. \end{aligned}$$


$$\begin{aligned} \eta (x + 2^{-\ell -1} A_k^{-1}) \ge \lambda _{1} + 2^{-\ell - 1} \cdot 2^{-\ell -1} A_k^{-1} > \lambda _2. \end{aligned}$$

Since \(\eta \) is continuous, there exists \(y \in T_{k, \ell +1}^{+}(\lambda _{2})\cap (x,x+ 2^{-\ell -1} A_k^{-1}]\). We say that y is “paired to x.” Notice that \(y - x < 2^{-\ell -1} A_k^{-1} \ge 2^{-\ell - 1}\) while \(\eta \le \lambda _1\) on \([x - 2^{-\ell -1} A_k^{-1}, x]\), so y is not paired to any other \(x \in T_{k, \ell }^{+}(\lambda _{1})\). Thus to each \(x \in T_{k, \ell }^{+}(\lambda _{1})\) we have paired a unique \(y \in T_{k, \ell +1}^{+}(\lambda _{2})\). Also,

$$\begin{aligned} |\eta '(x)|\zeta _{k}(x)-|\eta '(y)|\zeta _{k}(y) \le \Vert \eta ' \zeta _k\Vert _{{\mathcal {C}}^1({\mathbb {R}})} 2^{-\ell -1} A_k^{-1} \le 2^{-\ell - 1}. \end{aligned}$$

Now consider the difference

$$\begin{aligned} \sum _{x\in T_{k, \ell }^{+}(\lambda _{1})}|\eta '(x)|\zeta _{k}(x)-\sum _{y\in T_{k, \ell }^{+}(\lambda _{2})}|\eta '(y)|\zeta _{k}(y). \end{aligned}$$

Each \(x \in T_{k, \ell - 1}^+(\lambda _1)\) is paired to a unique \(y \in T_{k, \ell }^{+}(\lambda _{2})\), and the corresponding terms’ difference is at most \(2^{-\ell - 1}\). On the other hand, if \(x \in T_{k, \ell }^+(\lambda _1) \setminus T_{k, \ell - 1}^+(\lambda _1)\), we have

$$\begin{aligned} |\eta '(x)|\zeta _{k}(x) \le 2^{-\ell + 1} A_k. \end{aligned}$$

Decomposing (D.6) into these two cases, we obtain

$$\begin{aligned} \sum _{x\in T_{k, \ell }^{+}(\lambda _{1})}|\eta '(x)|\zeta _{k}(x)-\sum _{y\in T_{k, \ell }^{+}(\lambda _{2})}|\eta '(y)|\zeta _{k}(y)&\le \sum _{x \in T_{k, \ell }^+(\lambda _1)} (2^{-\ell - 1} + 2^{\ell + 1}A_k) \\ {}&\le 2^2 A_k 2^{-\ell } |T_{k, \ell }^+(\lambda _1)|. \end{aligned}$$

Now (D.4) implies

$$\begin{aligned} |T_{k, \ell }^+(\lambda _1)| \le \sum _{\ell ' \le \ell } |S_{k, \ell '}(\lambda _1)| \le 2^3 A_k^{1/2} 2^{\ell /2} \sum _{m \ge 0} 2^{-m/2} \le 2^5 A_k^{1/2} 2^{\ell /2}. \end{aligned}$$


$$\begin{aligned} \sum _{x\in T_{k, \ell }^{+}(\lambda _{1})}|\eta '(x)|\zeta _{k}(x)-\sum _{y\in T_{k, \ell }^{+}(\lambda _{2})}|\eta '(y)|\zeta _{k}(y)&\le 2^2 A_k 2^{-\ell } |T_{k, \ell }^+(\lambda _1)| \\&\le 2^7 A_k^{3/2} 2^{-\ell /2}. \end{aligned}$$

Symmetric arguments show that in fact

$$\begin{aligned} \Bigg |\sum _{x\in T_{k, \ell }^{\pm }(\lambda _{1})}|\eta '(x)|\zeta _{k}(x)-\sum _{y\in T_{k, \ell }^{\pm }(\lambda _{2})}|\eta '(y)|\zeta _{k}(y)\Bigg |\le 2^7 A_{k}^{3/2} 2^{-\ell /2}. \end{aligned}$$

Now consider \(g_k(\lambda _1) - g_k(\lambda _2)\). We divide (D.3) into terms in \(T_{k, \ell }^\pm (\lambda _i)\) or \(S_{k, \ell '}(\lambda _i)\) for \(\ell ' > \ell \) and \(i \in \{1, 2\}\). We pair terms in \(T_{k, \ell }^\pm (\lambda _1)\) with those in \(T_{k, \ell }^\pm (\lambda _2)\) to take advantage of the cancellation in (D.7). We treat the terms in \(S_{k, \ell '}(\lambda _i)\) as error and use (D.5). Then

$$\begin{aligned} |g_k(\lambda _1) - g_k(\lambda _2)| \le 2^8 A_k^{3/2} 2^{-\ell /2} + 2^5 A_k^{3/2} \sum _{\ell ' > \ell } 2^{-\ell '/2} \le 2^9 A_k^{3/2} 2^{-\ell /2} < \varepsilon . \end{aligned}$$

It follows that \(g_k\) is uniformly continuous.

We now wish to sum over k to conclude the same for g. To do so, we bound \(g_k\). Let \({\ell _k = -\log _2 A_k + 1}\). Then \(S_{k, \ell }(\lambda ) = \emptyset \) for all \(\lambda \in {\mathbb {R}}\) and \(\ell < \ell _k\). Hence (D.4) implies

$$\begin{aligned} g_k(\lambda )&= \sum _{\ell \ge \ell _k} \sum _{y \in S_{k, \ell }(\lambda )} |\eta '(y)| \zeta _k(y) \\&\le 2\Vert \zeta _k\Vert _{{\mathcal {C}}^0({\mathbb {R}})} \sum _{\ell \ge \ell _k} 2^{-\ell } |S_{k, \ell }(\lambda )| \\&\le 2^4 \Vert \zeta _k\Vert _{{\mathcal {C}}^0({\mathbb {R}})} A_k^{1/2} 2^{-\ell _k/2} \sum _{m \ge 0} 2^{-m/2}. \end{aligned}$$

Using the definition of \(\ell _k\), this yields

$$\begin{aligned} g_k(\lambda ) \le 2^5 \Vert \zeta _k\Vert _{{\mathcal {C}}^0({\mathbb {R}})} A_k. \end{aligned}$$

By hypothesis, there exists \(C \ge 1\) independent of k such that \(A_k \le C \langle k \rangle ^C\) for all \(k \in {\mathbb {Z}}\). But \(\zeta _k \le \zeta \) decays super-polynomially as \(|k| \rightarrow \infty \). It follows that \(\Vert g_k\Vert _{{\mathcal {C}}^0({\mathbb {R}})}\) itself decays super-polynomially in k. Therefore \(g=\sum \limits _{k\in {\mathbb {Z}}}g_{k}\) is an absolutely and uniformly convergent sum of uniformly continuous functions and hence is (uniformly) continuous.

\(\square \)

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Dunlap, A., Graham, C. & Ryzhik, L. Stationary Solutions to the Stochastic Burgers Equation on the Line. Commun. Math. Phys. 382, 875–949 (2021).

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