= Solution
The natural expectation is <orbital stability>, since a spatial translation of a <travelling wave> is another <travelling wave>. Thus convergence for ordinary perturbations, when it holds, is to a suitably translated front. It also matters which perturbations are allowed in the leading edge: the zero state ahead of this front is linearly unstable, and a sufficiently shallow positive tail can select a different speed. One should not assert <asymptotic stability> for arbitrary disturbances without a norm and a class of initial data. The requested <essential spectrum> calculation is a necessary part of such an analysis, not by itself a complete nonlinear stability proof.
In the moving coordinate $z=x-ct$, write $u(x,t)=w_c(z)+v(z,t)$. Expansion of the reaction term gives
$$
v_t=v_{zz}+cv_z+(k-2w_c)v-v^2.
$$
The first equation to study is its <linearization>,
$$
v_t=\mathcal Lv,\qquad \mathcal L=\partial_z^2+c\partial_z+k-2w_c(z),
$$
and the associated <eigenvalue> equation $\mathcal L\phi=\lambda\phi$. Differentiating the front equation gives $\mathcal Lw_c'=0$, the neutral translation mode whenever it belongs to the chosen function space. Ahead of the front the limiting coefficient is $k$, so in unweighted <L2 space> the limiting dispersion relation is $\lambda=-\xi^2+ic\xi+k$: its rightmost real part is positive. An exponential weight penalizes precisely these disturbances ahead of the front.
There is a small functional-analytic correction to the printed definition. Use the complete <exponentially weighted L2 space>
$$
X_\gamma=\{v:e^{\gamma z}v\in L^2(\mathbb R)\},\qquad
\|v\|_\gamma=\|e^{\gamma z}v\|_2.
$$
The additional requirement of unweighted <L2 space> membership in the printed set makes that set incomplete under its stated weighted norm. Indeed $v_N=\mathbf1_{[-N,0]}$ belongs to the intersection and is a weighted <Cauchy sequence>, but its weighted limit $\mathbf1_{(-\infty,0]}$ does not belong to unweighted <L2 space>. The usual <Hilbert space> for this spectral argument is the completion $X_\gamma$. We use that completion explicitly, rather than treating the intersection as a complete space.
Multiplication $T_\gamma v=e^{\gamma z}v$ is a unitary map from $X_\gamma$ onto <L2 space>. Take the domain $T_\gamma^{-1}H^2(\mathbb R)$ for $\mathcal L$. The transformed <closed operator> is
$$
\mathcal L_\gamma=T_\gamma\mathcal LT_\gamma^{-1}
=\partial_z^2+(c-2\gamma)\partial_z+\gamma^2-c\gamma+k-2w_c(z).
$$
This follows from $T_\gamma\partial_zT_\gamma^{-1}=\partial_z-\gamma$. The two limiting constant-coefficient dispersion relations, obtained by the <Fourier transform>, are
$$
\lambda_{\mathrm{ahead}}(\xi)=-\xi^2+i(c-2\gamma)\xi+\gamma^2-c\gamma+k,
$$
$$
\lambda_{\mathrm{behind}}(\xi)=-\xi^2+i(c-2\gamma)\xi+\gamma^2-c\gamma-k.
$$
In particular the necessary strict leading-edge condition is $\gamma^2-c\gamma+k<0$. For $c>2\sqrt{k}$ this means
$$
\frac{c-\sqrt{c^2-4k}}2<\gamma<\frac{c+\sqrt{c^2-4k}}2.
$$
A particularly convenient choice is
$$
\boxed{\gamma=c/2,\qquad k-c^2/4<0\quad\text{if }c>2\sqrt{k}.}
$$
With this choice there is no first derivative, and $\mathcal L_\gamma=\partial_z^2+k-c^2/4-2w_c(z)$ is a <self-adjoint differential operator> on $H^2(\mathbb R)$. This also avoids possible ambiguities between definitions of <essential spectrum> for nonnormal operators.
The relevant perturbation theorem is <essential-spectrum invariance under relatively compact perturbations>: if $A_0$ is a <closed operator>, $V:D(A_0)\to H$ is compact with respect to the <graph norm>, and $A_0+V$ is closed on the same domain, their Fredholm <essential spectrum> is the same. For <self-adjoint> operators this is the usual <essential spectrum> after removing isolated finite-multiplicity <eigenvalues>. To see why the theorem applies, regard $A_0-\lambda$ and $A_0+V-\lambda$ as bounded maps from the <graph norm> domain to $H$, and use <compact perturbation invariance of Fredholm operators> for each $\lambda$.
Define a step-potential comparison operator
$$
A_0=\partial_z^2+q_0(z),\qquad
q_0(z)=\begin{cases}-k-c^2/4,&z<0,\\k-c^2/4,&z\geq0.\end{cases}
$$
Both $A_0$ and $\mathcal L_\gamma$ are <self-adjoint> on $H^2(\mathbb R)$. Their difference is multiplication by
$$
V(z)=k-c^2/4-2w_c(z)-q_0(z),\qquad V(z)\longrightarrow0\quad(z\to\pm\infty).
$$
It is relatively compact, not necessarily compact as a multiplication map on all of <L2 space>. To prove the needed compactness, take a sequence bounded in $H^2(\mathbb R)$. On each bounded interval the <Rellich-Kondrachov compactness theorem> gives a subsequence converging in <L2 space>. A diagonal subsequence makes the products by $V$ converge locally. Outside a sufficiently large interval, $\|V\|_\infty$ is arbitrarily small and the sequence has a uniform <L2 norm> bound, so those tails are uniformly small. Thus the products converge globally in <L2 space>, proving graph-domain compactness.
Finally,
$$
\sigma_{\mathrm{ess}}(A_0)=(-\infty,k-c^2/4].
$$
Here is a direct verification. Its quadratic form is at most $(k-c^2/4)\|h\|_2^2$, so no <spectrum> lies above this endpoint. For every $\lambda\leq k-c^2/4$, choose $\xi$ with $\lambda=k-c^2/4-\xi^2$. Normalized, increasingly long cutoffs of $e^{i\xi z}$, translated far to the right, have disjoint supports and satisfy $\|(A_0-\lambda)h_N\|_2\to0$: differentiating the cutoff gives errors of orders $N^{-1}$ and $N^{-2}$. They converge weakly to zero. This is a singular sequence in the <Weyl criterion>, proving that every such $\lambda$ belongs to the <essential spectrum>. The relatively compact perturbation theorem therefore gives
$$
\boxed{\sigma_{\mathrm{ess}}(\mathcal L\text{ on }X_{c/2})=(-\infty,k-c^2/4].}
$$
This proves the <weighted essential spectrum of a Fisher travelling front> is strictly stable when $c>2\sqrt{k}$. In fact here the quadratic form gives more:
$$
\langle\mathcal L_{c/2}h,h\rangle
=-\|h'\|_2^2+\int(k-c^2/4-2w_c)|h|^2\,dz
\leq-(c^2/4-k)\|h\|_2^2.
$$
Since $w_c\geq0$, the full transformed <spectrum> lies below $-(c^2/4-k)$, and the evolution given by its <linearization> decays at least as $e^{-(c^2/4-k)t}$ in this weighted norm. This does not contradict translation invariance: the nonzero translation mode cannot lie in this weighted operator domain, since otherwise it would be a zero <eigenfunction>, contradicting the strictly negative quadratic-form bound. The weight restricts the permitted perturbations and excludes these neutral translations. Nor does a weighted <L2 norm> alone control the nonlinear remainder at $z\to-\infty$; a nonlinear theorem needs further boundedness or regularity assumptions.
At the minimal speed $c_*=2\sqrt{k}$, the best choice is $\gamma=\sqrt{k}$. Now $\gamma^2-c\gamma+k=(\gamma-\sqrt{k})^2$, and
$$
\boxed{\sigma_{\mathrm{ess}}(\mathcal L\text{ on }X_{\sqrt{k}})=(-\infty,0].}
$$
The <essential spectrum> has no positive part but touches zero; no positive exponential weight makes it strictly negative. Thus the critical front has no spectral gap and this argument gives no exponential decay rate there. \b[Strict weighted spectral stability holds above the minimal speed; at the minimal speed it is marginal.]
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