We define $\gamma_\lambda(y) = \exp(-y^2/\lambda)$ and for all $x \in \mathbf{R}$, $\tau_x(f)(y) = f(y-x)$. For all $f, g \in \mathcal{E}$, $(f \mid g) = \int_{-\infty}^{+\infty} f(y)g(y)\,\mathrm{d}y$. (a) Let $a > 0$. Show that there exists $c \geq 0$ such that for all $x \in \mathbf{R}$ we have $$\int_{-\infty}^{+\infty} \exp\left(-\frac{(y-x)^2}{\lambda}\right) \exp\left(-\frac{y^2}{a}\right) \mathrm{d}y = c \exp\left(-\frac{x^2}{a+\lambda}\right).$$ Hint: One may show the equality $$\frac{(y-x)^2}{\lambda} + \frac{y^2}{a} = \frac{a+\lambda}{a\lambda}\left(y - \frac{ax}{a+\lambda}\right)^2 + \frac{x^2}{a+\lambda}.$$ (b) Let $g \in \mathcal{E}$. We consider $C(g) : \mathbf{R} \rightarrow \mathbf{R}$ defined for all $x \in \mathbf{R}$ by $$C(g)(x) = \left(\tau_x(\gamma_\lambda) \mid g\right)$$ Show that $C(g) \in \mathcal{E}$. (c) Show that $C : \mathcal{E} \rightarrow \mathcal{E}$ defines an endomorphism of $\mathcal{E}$.
We define $\gamma_\lambda(y) = \exp(-y^2/\lambda)$ and for all $x \in \mathbf{R}$, $\tau_x(f)(y) = f(y-x)$. For all $f, g \in \mathcal{E}$, $(f \mid g) = \int_{-\infty}^{+\infty} f(y)g(y)\,\mathrm{d}y$.
(a) Let $a > 0$. Show that there exists $c \geq 0$ such that for all $x \in \mathbf{R}$ we have
$$\int_{-\infty}^{+\infty} \exp\left(-\frac{(y-x)^2}{\lambda}\right) \exp\left(-\frac{y^2}{a}\right) \mathrm{d}y = c \exp\left(-\frac{x^2}{a+\lambda}\right).$$
Hint: One may show the equality
$$\frac{(y-x)^2}{\lambda} + \frac{y^2}{a} = \frac{a+\lambda}{a\lambda}\left(y - \frac{ax}{a+\lambda}\right)^2 + \frac{x^2}{a+\lambda}.$$
(b) Let $g \in \mathcal{E}$. We consider $C(g) : \mathbf{R} \rightarrow \mathbf{R}$ defined for all $x \in \mathbf{R}$ by
$$C(g)(x) = \left(\tau_x(\gamma_\lambda) \mid g\right)$$
Show that $C(g) \in \mathcal{E}$.
(c) Show that $C : \mathcal{E} \rightarrow \mathcal{E}$ defines an endomorphism of $\mathcal{E}$.