Uniform convergence of $f_{n}\left(x\right)=nxe^{-n^2x^2}$ on $0\le x<\infty $ and $ 2\le x<\infty $

Question

I need to check if Uniform convergence of $f_{n}(x)=nxe^{-n^2x^2}$ on $0\le x<\infty $ and $ 2\le x<\infty $

my suggest answer for $0\le x<\infty $ :

$$f(x)=\lim _{n\to \infty }(nxe^{-n^2x^2})=\lim \:_{n\to \:\infty \:}(\frac{nx}{e^{n^2x^2}})=0\:$$

let $$g_n(x)=nxe^{-n^2x^2}$$

$$g'_n(x)=e^{-n^2x^2}n(1-2n^2x^2)=0 \implies x=\frac{1}{\sqrt{2}n}$$

$$max_{x \in [0,\infty)} g_n(x)=g(\frac{1}{\sqrt{2}n})=\frac{1}{\sqrt{2e}}\neq0$$

so we get that in $[0,\infty) the $ convergence not uniform

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Is this answer correct? and any idea How to check when $ 2\le x<\infty $?

thanks

Answer 1

It follows from your computation of $g_n'$ that $g_n$ is decreasing in $[2,+\infty)$. So$$\bigl(\forall x\in[2,+\infty)\bigr):g_n(x)\leqslant g_n(2)=\frac{2n}{e^{4n^2}}\to0.$$

Answer 2

Short answer: The sequence $(f_n)$ converges uniformly on $[2,\infty)$ but not on $[0,\infty).$

Long answer: Since the maximum of $f_n$ approaches to $0$ for each natural number $n,$ there does not exist $N$ such that for all $n\geq N,$ $|f_n(x)|<\varepsilon.$

Formal proof: Let $\varepsilon=\frac{1}{10}.$ Fix a natural number $N.$ Let $x = \frac{1}{\sqrt{2}N}$ and $n=N.$ Then $$|f_n(x) - f(x)| = \frac{1}{\sqrt{2e}}>\frac{1}{10}.$$

While on $[2,\infty),$ we can choose large $n$ so that the maximum of $(f_n)$ falls on the left of $n$ for all $n\geq N$. Therefore, we can have $|f_n(x)|<\varepsilon.$