Supremum $\sup_{x\in (0,+\infty)}|F(x,t)|=\sup_{x\in (0,+\infty)}\dfrac{e^{-t^2}}{x+|t|}$

Question

Let :
\begin{aligned}F \colon (0,+\infty)\times \mathbb{R} &\longrightarrow \mathbb{R} \\(x,t) &\longmapsto F(x,t)=\dfrac{e^{-t^2}}{x+|t|}.\end{aligned}

• How they do to find $\displaystyle \sup_{x\in (0,+\infty)}|F(x,t)|$

$$ \fbox{ $\sup_{x\in (0,+\infty)}|F(x,t)|=\sup_{x\in (0,+\infty)}\dfrac{e^{-t^2}}{x+|t|}=\dfrac{e^{-t^2}}{|t|} $} $$

Indeed,

$\forall t\in \mathbb{R}\quad F(.,t)=\dfrac{e^{-t^2}}{x+|t|}$ $$\frac{\partial F}{\partial x}(x,t)=\dfrac{-e^{-t^2}}{(x+|t|)^2} $$

I'm stuck here

Answer 1

For any fixed $t$, one may see it as a function of a single variable $x$: $$ f(x)=\dfrac{e^{-t^2}}{x+|t|}, \quad x>0, $$ giving $$ f'(x)=-\dfrac{e^{-t^2}}{(x+|t|)^2}<0, $$ thus $f$ is decreasing over $(0,\infty)$, its supremum is obtained as $x \to 0^+$: $$ \lim_{x \to 0^+}f(x)=\lim_{x \to 0^+}\dfrac{e^{-t^2}}{x+|t|}=\dfrac{e^{-t^2}}{|t|}. $$

Answer 2

The supremum of this value is achieved when the denominator has the infimum value which is achieved for $x=0$.