For the proof of the main result, we need the following lemmas.

**Lemma 3.1** *Let* $\rho \in {\mathcal{U}}_{T}$ *and* $u\in {C}^{1}(\mathbb{R},X)\cap {E}^{0}(X)$. *If the derivative* ${u}^{\prime}$ *is uniformly continuous*, *then* ${u}^{\prime}\in {E}^{0}(X)$, *thus* $u\in {E}^{1}(X)$.

**Remark 3.2** Lemma 3.1 is well know for $A{P}^{0}(X)$, $A{A}^{0}(X)$. In the scalar case, for $PA{P}^{0}(X)$, this result is proved in [12], Corollary 5.6, p.59.

*Proof* Consider the function ${g}_{n}:\mathbb{R}\to X$ defined by ${g}_{n}(t):=n(f(t+\frac{1}{n})-f(t))$ for $n\in \mathbb{N}\setminus \{0\}$. Since ${E}^{0}(X)$ is a translation invariant vectorial space, then ${g}_{n}\in {E}^{0}(X)$. The equality ${g}_{n}(t)-{f}^{\prime}(t)=n{\int}_{0}^{\frac{1}{n}}({f}^{\prime}(t+s)-{f}^{\prime}(t))\phantom{\rule{0.2em}{0ex}}ds$ shows that the uniform continuity of ${f}^{\prime}$ implies that the sequence ${({g}_{n})}_{n}$ with values in the Banach space ${E}^{0}(X)$ converges uniformly to ${f}^{\prime}$ on ℝ. Then ${f}^{\prime}\in {E}^{0}(X)$, and from the definition of ${E}^{1}(X)$, we obtain $f\in {E}^{1}(X)$. □

**Lemma 3.3** *Let* $\rho \in {\mathcal{U}}_{T}$ *and* $A\in \mathcal{L}(X,X)$.

*If the spectrum* $\sigma (A)$ *of* *A* *does not intersect the imaginary axis*,

*then for all* $h\in {E}^{0}(X)$,

*there exists a unique solution in* ${E}^{0}(X)$ *of the differential equation* ${u}^{\prime}(t)=Au(t)+h(t).$

(3.1)

*Moreover*, *the solution* *u* *is in* ${E}^{1}(X)$.

*Proof* Applying Theorem 4.1, p.81 in [

20] (or Theorem 4 in [

21]), Equation (

3.1) admits a unique bounded solution on ℝ which is given by the formula

$u(t)={\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}G(t-s)h(s)\phantom{\rule{0.2em}{0ex}}ds,$

(3.2)

where *G* is the principal Green function for Equation (3.1). The Green function $G:\mathbb{R}\to \mathcal{L}(X,X)$ is continuous on $\mathbb{R}-\{0\}$, and there exist $M\ge 1$ and $\omega >0$ such that $\parallel G(t)\parallel \le Mexp(-\omega |t|)$ for all $t\in \mathbb{R}$.

Now we prove that the bounded solution *u* defined by (3.2) belongs to ${E}^{0}(X)$.

When ${E}^{0}(X)=A{P}^{0}(X)$ (respectively ${E}^{0}(X)=A{A}^{0}(X)$), this result is a straightforward consequence of Theorem 3.8 in [15] (respectively Theorem 3.9 in [16]) and when ${E}^{0}(X)={C}_{0}(X)$, this result is proved in [21], Proposition 3. Then we deduce the result for ${E}^{0}(X)=AA{P}^{0}(X)$ (respectively ${E}^{0}(X)=AA{A}^{0}(X)$).

Note that the case of the pseudo almost periodic (respectively pseudo almost automorphic) functions is a special case of the weighted pseudo almost periodic (respectively weighted pseudo almost automorphic) functions by taking $\rho (t):=1$ for all $t\in \mathbb{R}$; remark that the associated measure is exactly the Lebesgue measure.

And so it suffices to prove the cases of weighted pseudo almost periodic functions and of weighted pseudo almost automorphic functions.

When ${E}^{0}(X)=WPA{P}^{0}(X,\rho )$ (respectively ${E}^{0}(X)=WPA{A}^{0}(X,\rho )$), this result is a straightforward consequence of Theorem 3.8 in [15] (respectively Theorem 3.9 in [16]). Consequently, $u\in {E}^{0}(X)$, and from the definition of ${E}^{0}(X)$, we deduce that $[t\mapsto Au(t)]\in {E}^{0}(X)$. Since *u* satisfies Equation (3.1), then ${u}^{\prime}\in {E}^{0}(X)$, and from the definition of ${E}^{1}(X)$, we obtain $u\in {E}^{1}(X)$. □

**Lemma 3.4** *Let* *X* *and* *Y* *be two finite*-*dimensional Banach spaces*, *and let* $\varphi :X\to Y$ *be a continuous mapping*. *Then the Nemytskii operator* ${N}_{\varphi}:{E}^{0}(X)\to {E}^{0}(Y)$, *defined by* ${N}_{\varphi}(u):=[t\mapsto \varphi (u(t))]$, *is continuous*.

**Remark 3.5** Contrary to the asymptotically almost periodic case and, in particular, for the almost periodic case, when the dimension of the Banach spaces *X* and *Y* is infinite, Lemma 3.4 does not hold for the pseudo almost periodic case, and thus for the weighted pseudo almost periodic case, without additional assumptions. This is due to the fact that the range of a pseudo almost periodic function is only bounded, but not relatively compact, contrary to the asymptotically almost periodic case. This last observation still holds when the word almost periodic is replaced by almost automorphic.

*Proof* When ${E}^{0}(X)=AA{P}^{0}(X)$ and ${E}^{0}(Y)=AA{P}^{0}(Y)$, replacing ${\mathbb{R}}_{+}$ by ℝ, this result is a variation of Theorem 8.4 in [22].

When ${E}^{0}(X)=AA{A}^{0}(X)$ and ${E}^{0}(Y)=AA{A}^{0}(Y)$, replacing ${\mathbb{R}}_{+}$ by ℝ, the inclusion ${N}_{\varphi}(AA{A}^{0}(X))\subset AA{A}^{0}(Y)$ is a variation of Theorem 2.15 in [9]. Moreover, using Lemma 1 in [23], we know that ${N}_{\varphi}:B{C}^{0}(\mathbb{R},X)\to B{C}^{0}(\mathbb{R},Y)$ is continuous, and so its restriction to $AA{A}^{0}(X)$ is also continuous.

When ${E}^{0}(X)=PA{P}^{0}(X)$ and ${E}^{0}(Y)=PA{P}^{0}(Y)$ (respectively ${E}^{0}(X)=PA{A}^{0}(X)$ and ${E}^{0}(Y)=PA{A}^{0}(Y)$), this result is a straightforward consequence of Theorem 4.1 (respectively Theorem 4.2) in [24].

When ${E}^{0}(X)=WPA{P}^{0}(X,\rho )$ and ${E}^{0}(Y)=WPA{P}^{0}(Y,\rho )$ (respectively ${E}^{0}(X)=WPA{A}^{0}(X,\rho )$ and ${E}^{0}(Y)=WPA{A}^{0}(Y,\rho )$), using Corollary 4.12 in [15] (respectively Corollary 5.10 in [16]), we know that ${N}_{\varphi}(WPA{P}^{0}(X,\rho ))\subset WPA{P}^{0}(Y,\rho )$ (respectively ${N}_{\varphi}(WPA{A}^{0}(X,\rho ))\subset WPA{P}^{0}(Y,\rho )$). Moreover, using Lemma 1 in [23], we know that ${N}_{\varphi}:B{C}^{0}(\mathbb{R},X)\to B{C}^{0}(\mathbb{R},Y)$ is continuous, and so its restriction to $WPA{P}^{0}(X,\rho )$ (respectively $WPA{A}^{0}(X,\rho )$) is also continuous. □

**Lemma 3.6** *Let* *X* *and* *Y* *be two finite*-*dimensional Banach spaces*, *and let* $\varphi :X\to Y$ *be a continuously Fréchet*-*differentiable mapping*. *Then the Nemytskii operator* ${N}_{\varphi}:{E}^{0}(X)\to {E}^{0}(Y)$ *is continuously Fréchet*-*differentiable on* ${E}^{0}(X)$, *and we have* $D{N}_{\varphi}(u)v=[t\mapsto D\varphi (u(t))v(t)]$ *for all* $u,v\in {E}^{0}(X)$.

*Proof* When ${E}^{0}(X)=AA{P}^{0}(X)$ and ${E}^{0}(Y)=AA{P}^{0}(Y)$, replacing ${\mathbb{R}}_{+}$ by ℝ, this result is a variation of Theorem 8.5 in [22].

When ${E}^{0}(X)=AA{A}^{0}(X)$ and ${E}^{0}(Y)=AA{A}^{0}(Y)$, using Lemma 1 in [23], we know that ${N}_{\varphi}:B{C}^{0}(\mathbb{R},X)\to B{C}^{0}(\mathbb{R},Y)$ is of class ${C}^{1}$ and that we have $D{N}_{\varphi}(u)h=[t\mapsto D\varphi (u(t))\cdot h(t)]$ when $u,h\in B{C}^{0}(\mathbb{R},X)$. Now, using Theorem 2.15 in [9], we know that ${N}_{\varphi}(AA{A}^{0}(X))\subset AA{A}^{0}(Y)$ and that $D\varphi \circ u\in AA{A}^{0}(\mathcal{L}(X,Y))$ when $u\in AA{A}^{0}(X)$. And so ${N}_{\varphi}\in {C}^{1}(AA{A}^{0}(X),AA{A}^{0}(Y))$ and the announced formula for its Fréchet-differential is proven.

As in the proof of Lemma 3.3, the case of $PA{P}^{0}(X)$ (respectively $PA{A}^{0}(X)$) is a corollary of the case $WPA{P}^{0}(X,\rho )$ (respectively $WPA{A}^{0}(X,\rho )$). And so it suffices to prove the cases of the weighted pseudo almost periodic functions and of the weighted pseudo almost automorphic functions.

To prove the result in the case where ${E}^{0}(X)=WPA{P}^{0}(X,\rho )$ and ${E}^{0}(Y)=WPA{P}^{0}(Y,\rho )$ (respectively ${E}^{0}(X)=WPA{A}^{0}(X,\rho )$ and ${E}^{0}(Y)=WPA{A}^{0}(Y,\rho )$), note that, using Lemma 1 in [23], we know that ${N}_{\varphi}:B{C}^{0}(\mathbb{R},X)\to B{C}^{0}(\mathbb{R},Y)$ is of class ${C}^{1}$ and that we have $D{N}_{\varphi}(u)h=[t\mapsto D\varphi (u(t))\cdot h(t)]$ when $u,h\in B{C}^{0}(\mathbb{R},X)$. Now, using Corollary 4.12 in [15] (respectively Corollary 5.10 in [16]), we know that ${N}_{\varphi}(WPA{P}^{0}(X,\rho ))\subset WPA{P}^{0}(Y,\rho )$ (respectively ${N}_{\varphi}(WPA{A}^{0}(X,\rho ))\subset WPA{A}^{0}(Y,\rho )$) and that $D\varphi \circ u\in WPA{P}^{0}(\mathcal{L}(X,Y),\rho )$ (respectively $D\varphi \circ u\in WPA{A}^{0}(\mathcal{L}(X,Y),\rho )$) when $u\in WPA{P}^{0}(X,\rho )$ (respectively $WPA{A}^{0}(X,\rho )$).

Consequently, we obtain ${N}_{\varphi}\in {C}^{1}(WPA{P}^{0}(X,\rho ),WPA{P}^{0}(Y,\rho ))$ (respectively ${N}_{\varphi}\in {C}^{1}(WPA{A}^{0}(X,\rho ),WPA{A}^{0}(Y,\rho ))$) and the announced formula for its Fréchet-differential is proven. □

**Lemma 3.7** *Let* $\rho \in {\mathcal{U}}_{T}$, $p\in P$ *and* ${e}_{p}\in {E}^{0}(\mathbb{R})$. *If* *x* *is a solution of* ($\mathcal{E},p$) *in* ${E}^{0}(\mathbb{R})$, *that is*, $x\in {C}^{2}(\mathbb{R},\mathbb{R})\cap {E}^{0}(\mathbb{R})$ *and* *x* *satisfies* ($\mathcal{E},p$), *then* $x\in {E}^{2}(\mathbb{R})$.

*Proof* Lemma 3.2 in [4] asserts that if *x* is a solution of ($\mathcal{E},p$) in $B{C}^{0}(\mathbb{R},\mathbb{R})$, then $x\in B{C}^{2}(\mathbb{R},\mathbb{R})$, therefore the derivative ${x}^{\prime}$ is uniformly continuous, and by help of Lemma 3.1, we obtain $x\in {E}^{1}(\mathbb{R})$. By using Lemma 3.4, the functions $[t\mapsto f(x(t),p)]$ and $[t\mapsto g(x(t),p)]$ are in ${E}^{0}(\mathbb{R})$. Applying again Lemma 3.4 to $[t\mapsto (f(x(t),p),{x}^{\prime}(t))]\in {E}^{0}({\mathbb{R}}^{2})$ and using the continuous function $\varphi :{\mathbb{R}}^{2}\to \mathbb{R}$ defined by $\varphi (r,s):=r\cdot s$, we obtain that $[t\mapsto f(x(t),p)\cdot {x}^{\prime}(t)]\in {E}^{0}(\mathbb{R})$. Since ${x}^{\u2033}(t)={e}_{p}(t)-f(x(t),p)\cdot {x}^{\prime}(t)-g(x(t),p)$, then ${x}^{\u2033}\in {E}^{0}(\mathbb{R})$, and from the definition of ${E}^{2}(\mathbb{R})$, we obtain that $x\in {E}^{2}(\mathbb{R})$. □