Let $Q: \mathbb{R} \rightarrow \mathbb{R}$ be a $C_c^{\infty}$ function, i.e. it is smooth and has c ompact support. We assume $Q$ is even, i.e. $Q(x)=Q(-x)$. We assume $Q$ is non-trivial,(i.e. $Q$ does not equal to zero ever ywhere).

Let $T_1(x):=x Q(x)$, and let $T_2(x)=x^2 Q(x)$. Let $T_3:=e^{-x^2}\left(1+x^{2024}\right)$

We also introduce the following notation. For any $f: \mathbb{R} \rightarrow \mathbb{R}, \lambda>0, \alpha \in \mathbb{R}$, we define
$$
f_{\lambda, \alpha}:=\frac{1}{\lambda^{1 / 2}} f\left(\frac{x-\alpha}{\lambda}\right)
$$

We claim: There exists $\delta>0, \varepsilon>0$, so that for any $c \in \mathbb{R}$ wit $\mathrm{h}|c| < \delta$, one can find unique $\lambda, \alpha$ such that the following $\mathrm{h}$ old
(1) $|\lambda-1|+|\alpha| < \varepsilon$.
(2) $\left\langle Q_{\lambda, \alpha}-Q-c T_3, T_1\right\rangle=0$.
(3) $\left\langle Q_{\lambda, \alpha}-Q-c T_3, T_2\right\rangle=0$.

(Here, for any two functions $f_1, f_2$, we define $\left.\left\langle f_1, f_2\right\rangle:=\int f_1(x) f_2(x) \mathrm{d} x\right)$.

Is the above claim correct? Prove your conclusion.

Recall for every $f \in L^2\left(\mathbb{R}^3\right)$, one has that $g(x):=(-\Delta+1)^{-1} f$ is a well-defined $L^2\left(\mathbb{R}^3\right)$ function. A nd one may compute $g$ by solving
$$
(-\Delta+1) g=f
$$
(Recall $\Delta$ in $\mathbb{R}^3$ is defined as $\Delta:=\sum_{i=1}^3 \partial_i^2$, also recall one may also define $(-\Delta+1)^{-1}$ by Fourier theory.)

Now, let $V(x):=e^{-|x|^2}, x \in \mathbb{R}^3$. Prove that the operator $T:=I+(-\Delta+1)^{-1} V$ is invertible in $L^2$.
(Here, $T f:=f+(-\Delta+1)^{-1}(V f)$.)

Let $\psi(\xi) \in C_c^{\infty}(\mathbb{R})$ be smooth and has compact support. Let $\psi(\xi)=0, \forall|\xi| \geq 1$. Let $f_1(\xi), f_2(\xi) \in C_c^{\infty}(\mathbb{R})$, i.e. $f_1, f_2$ ar e smooth and have compact support. Let $u_i: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{C}$, $i=1,2$, be defined as
$$
\begin{gathered}
u_1\left(x_1, x_2\right):=\int_{\mathbb{R}} \psi(\xi) f_1(\xi) e^{i \xi x_1} e^{i \xi^2 x_2} \mathrm{~d} \xi, \\
u_2\left(x_1, x_2\right):=\int_{\mathbb{R}} \psi(\eta-10) f_2(\eta) e^{i \eta x_1} e^{i \eta^2 x_2} \mathrm{~d} \eta .
\end{gathered}
$$

Prove there exists a constant $C$, which may depend on $\psi$, but does not depend on $f_1, f_2$, so that
$$
\left\|u_1 u_2\right\|_{L^2\left(\mathbb{R}^2\right)} \leq C\left\|f_1\right\|_{L^2(\mathbb{R})}\left\|f_2\right\|_{L^2(\mathbb{R})} .
$$
(Hint: One may try to use Plancherel Theorem. It may be usef ul to observe that if one let $H(\xi, \eta)=f_1(\xi) f_2(\eta)$, then $\|H\|_{L^2\left(\mathbb{R}^2\right)}$ are also bounded by $\left\|f_1\right\|_{L^2(\mathbb{R})}\left\|f_2\right\|_{L^2(\mathbb{R})}$

Consider the heat equation in $\mathbb{R}^2$. Let $u=u(t, x)$ is s solutio $\mathrm{n}$ to
$$
\left\{\begin{array}{l}
\frac{\partial u}{\partial t}-\Delta u=0 \\
\left.u\right|_{t=0}=u_0 \in L^2
\end{array}\right.
$$

Then there exists a universal constant $C$ such that
$$
\int_0^{\infty}\|u(t)\|_{L^2}^2 \mathrm{~d} t \leq C\left\|u_0\right\|_{L^2}^2
$$

Consider the Fourier transform. Let
$$
Q(g, f)(x):=\int_{\mathbb{R}^N} \int_{S^{N-1}} B\left(|x-y|, \frac{x-y}{|x-y|} \cdot \sigma\right) g\left(y^{\prime}\right) f\left(x^{\prime}\right) \mathrm{d} \sigma \mathrm{d} y
$$
where $B$ is a given two variable function, $S^{N-1}$ stands for the unit sphere in $\mathbb{R}^N$ and
$$
x^{\prime}:=\frac{x+y}{2}+\frac{|x-y| \sigma}{2} ; \quad y^{\prime}:=\frac{x+y}{2}-\frac{|x-y| \sigma}{2} .
$$

Then
$$
\widehat{Q(g, f)}(\xi)=(2 \pi)^{-N / 2} \int_{\mathbb{R}^N \times S^{N-1}} \hat{B}\left(|\eta|, \frac{\xi}{|\xi|} \cdot \sigma\right) \hat{g}\left(\xi^{-}+\eta\right) \hat{f}\left(\xi^{+}-\eta\right) \mathrm{d} \sigma \mathrm{d} \eta,
$$
where $\hat{B}(|\eta|, t):=\int_{\mathbb{R}^N} B(|q|, t) e^{-i q \cdot \eta} \mathrm{d} q, \xi^{ \pm}:=\frac{\xi \pm|\xi| \sigma}{2}$.