We denote for $x \in \mathbb{R}_{+}$ and $m \in \mathbb{R}$ $$T_{m}(x) = \int_{0}^{\infty} t^{m} e^{-\left(t^{2}+x/t\right)} dt$$
a) Show that for $x \in \mathbb{R}_{+}^{*}$, $$T_{-1}(x) \leq \int_{0}^{1} e^{-1/u^{2}} \frac{du}{u} + \int_{1}^{\infty} e^{-xu} \frac{du}{u}$$ Deduce that $T_{-1}(x) \leq 2$ for $x \geq 1$ and that $$T_{-1}(x) \leq 2 + \int_{x}^{1} e^{-w} \frac{dw}{w} \leq 2 - \ln x$$ if $0 < x \leq 1$.
b) Let $L \in [0,1]$, and $\rho \in C([0,L])$. We set $$[F(\rho)](x) = \int_{0}^{L} \rho(y) T_{-1}(|x-y|) dy$$ Show that $[F(\rho)](x)$ is well-defined for $x \in [0,L]$ and that $$\|F(\rho)\|_{\infty} \leq (4L + 2L|\ln L|) \|\rho\|_{\infty}.$$