139 $T$ not stable implies not BIBO stable

has a pole at $i\nu$, and hence $\widehat{Y}(s)=T(s)\widehat{U}(s)$ has a double (or triple, …) pole at $i\nu$. But $Y(t)$ is bounded, so $|Y(t)|\le M$ for some $M$ and all $t$, so

$$|\widehat{Y}(s)|=\left|{\int }_0^{\mathrm{\infty }}{e}^{-st}Y(t)𝑑t\right|$$

$$\le {\int }_0^{\mathrm{\infty }}{e}^{-t\mathrm{\Re }s}Mdt|$$

$$\le \frac{M}{\mathrm{\Re }s}.$$

Now consider $s$ with $\mathrm{\Re }s>0$ and $s\to i\nu$. Now $T(s)\widehat{U}(s)$ diverges like $1/{(s-i\nu )}^2$ or $1/{(s-i\nu )}^3$ etc.; whereas $\widehat{Y}(s)$ can only diverge like $M/\mathrm{\Re }s$ at worst. This contradicts the identity $\widehat{Y}(s)=T(s)\widehat{U}(s)$.