Does $f_n(x)=nxe^{-nx}\sin(x)$ converges uniformly on $\mathbb{R}^{+}$?

Question

Define the formula, $$\forall n\in\mathbb{N},\forall x\in\mathbb{R}^{+},\quad f_n(x)=nxe^{-nx}\sin(x)$$

Does the sequence $f_n$ converges uniformly on $\mathbb{R}^{+}$, as $n \to \infty$.

Let's check first the

Pointwise convergence:

$$f_{n}(0)=0,\quad f_n(0) \xrightarrow[n \to +\infty]{} 0$$

$$\forall x\in\mathbb{R}^{+},\quad f_{n}(x)\xrightarrow[n \to +\infty]{} 0$$

then $(f_n)_n\in\mathbb{N} $ does converge pointwise on $\mathbb{R}^+$ to the zero function

Converges uniformly:

we would like to check if $$\sup_{x\in\mathbb{R}^{+}}|f_n(x)|\underset{n \to +\infty}{\overset{}{\longrightarrow}} 0\iff \forall \epsilon >0\ \exists\ N\in\mathbb{N}\ \forall n\geq N\implies \forall x\in\mathbb{R}_{+}\quad |f_{n}(x)|\leq \epsilon$$ in order to confirm the convergens uniformly on $\mathbb{R}^{+}$

Since $$f_{n}'(x)= -n^2 x -e^{-n x}\sin(x)+n e^{-n x} sin(x)+n x e^{-n x} \cos(x)$$ then i will study the following function:

$$ h(x)=xe^{-x}$$

\begin{array}{c|ccccccc} x & 0 & & 0 & & +\infty \\ \hline h'(x) & & + & 0 & - & \\ \hline h(x) & 0 & \nearrow & \dfrac{1}{e} & \searrow & 1 & \end{array}

$$\sup_{x\in\mathbb{R}^{+}}h(x)=\dfrac{1}{e} $$ Let $\epsilon >0$

• I'm stuck here and i'm intersted in other ways to find $$\sup_{x\in\mathbb{R}^{+}}|f_n(x)|$$
• I'll be gratefull if someone could give detailled answer

Answer 1

Hint: The function is bounded in absolute value by $g_n(x)=nx^2 e^{-nx}$. It suffices to show that $g_n$ tends uniformly to zero (hint2: find its maximum) in order to conclude the same for $f_n$.

Added: Finding the extremum: $g_n'(x)=(2nx-n^2x^2)e^{-nx}=nx(2-nx)e^{-nx}$ which vanishes for $x=2/n$ (and for $x=0$). Checking monotonicity the first must be a maximum. So

$$ 0\leq \sup_{x\geq 0} g_n(x) \leq g_n\left( 2/n \right)=\frac{4e^{-2}}{n} $$ which goes to zero as $n\rightarrow +\infty$. And $|f_n(x)|\leq g_n(x)$.

Answer 2

We have that, as $n\to \infty$, $$\sup_{x\geq 0}|f_n(x)|\leq \sup_{x\geq 0}\left(nxe^{-nx}\cdot x\right)=\frac{1}{n}\cdot \sup_{t\geq 0}(t^2e^{-t})\leq \frac{M}{n}\to 0$$ where we used the fact that the function $t^2e^{-t}$ is bounded in $[0,+\infty)$.

Answer 3

For every positive $x$, $|\sin x|\le x$ and Maclaurin formula on the exponential yields $e^{nx}\ge1+nx+\frac12n^2x^2\ge\frac12n^2x^2$ hence $$\left| f_n(x)\right| =\left | \frac{nx\sin x}{e^{nx}} \right |\le \left | \frac{nx\cdot x}{\frac12n^{2}x^{2}} \right |=\frac{2}{n}$$