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Consider the sequence of functions $\displaystyle f_{n}:\mathbb{R}\to \mathbb{R}, \: f_{n}(x)=\sum_{k=1}^{n}\frac{k}{k^2x^2+2n^2}$. I want to determine that $\left\{f_{n} \right\}$ converges uniformly. I know that $\left\{f_{n} \right\}$ converges to $$f(x)= \begin{cases} \displaystyle \int_{0}^{1} \frac{t}{x^2t^2+2}dt, & \text{if }\; x\neq 0, \\ \displaystyle \frac{1}{4},& \text{if }\; x=0, \\ \end{cases}$$ but I can't determine uniform convergence. Any help please.

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    $\begingroup$ Note that the integral formula for $f(x)$ gives $1/4$ when $x=0$, so you do not need to distinguish two cases. $\endgroup$
    – Gary
    Jun 23 at 9:18

1 Answer 1

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Let $$F(x,t)=\frac t{x^2t^2+2},\qquad x\in\mathbb R, \ t\in[0,1].$$ Then $f_n(x)=\frac1n\sum_{k=1}^nF\left(x,\frac kn\right)$ and thus $$\lim_{n\to\infty}f_n(x)=\int_0^1F(x,t)\,dt=:f(x).$$ We claim that $$\sup_{x\in\mathbb R}|f_n(x)-f(x)|\leq\frac1{4n}.\tag{1}$$ As a result, we see that $f_n$ converges uniformly to $f$. Now we prove $(1)$. \begin{align*} |f_n(x)-f(x)|&=\left|\frac1n\sum_{k=1}^nF\left(x,\frac kn\right)-\int_0^1F(x,t)\,dt\right|\\ &=\left|\sum_{k=1}^n\int_{\frac{k-1}n}^\frac kn\left[F\left(x,\frac kn\right)-F(x,t)\right]\,dt\right|\\ &\leq \sum_{k=1}^n\int_{\frac{k-1}n}^\frac kn\left|F\left(x,\frac kn\right)-F(x,t)\right|\,dt. \end{align*} We use mean value theorem to estimate $\left|F\left(x,\frac kn\right)-F(x,t)\right|$. We calculate that $$\frac{\partial F}{\partial t}(x,t)=\frac{2-x^2t^2}{(x^2t^2+2)^2}=\frac4{(x^2t^2+2)^2}-\frac1{x^2t^2+2},\qquad x\in\mathbb R, t\in[0,1].$$ Consider $\phi(s)=4s^2-s$ for $s\in(0,1/2]$, then $|\phi(s)|\leq \max\{|\phi(1/8)|,|\phi(1/2)|\}=\phi(1/2)=\frac12$. Hence $$\left|\frac{\partial F}{\partial t}(x,t)\right|\leq\frac12,\qquad x\in\mathbb R, \ t\in[0,1],$$ and thus $\left|F\left(x,\frac kn\right)-F(x,t)\right|\leq \frac12\left|\frac kn-t\right|$ by mean value theorem. Therefore, \begin{align*} |f_n(x)-f(x)|&\leq\sum_{k=1}^n\int_{\frac{k-1}n}^\frac kn\left|F\left(x,\frac kn\right)-F(x,t)\right|\,dt\\ &\leq\frac12 \sum_{k=1}^n\int_{\frac{k-1}n}^\frac kn\left|\frac kn-t\right|\,dt\\ &=\frac14\sum_{k=1}^n\frac1{n^2}=\frac1{4n}. \end{align*} This gives the inequality $(1)$ and thus completes the proof of the uniform convergence.

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  • $\begingroup$ Wow! Thank you very much! $\endgroup$
    – purecj
    Jun 24 at 12:26
  • $\begingroup$ @purecj You’re welcome! $\endgroup$
    – Feng
    Jun 24 at 13:26

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