Uniform convergence and rigor proof of pointwise convergence of $f_n(x)=n\left(\sqrt{x+\frac1n}- \sqrt{x}\right)$ for all $x \in (0,\infty)$.

Question

Let $(f_n)$ be a sequence of functions where $f_n:(0,\infty) \to \Bbb R$ defined by $f_n(x)=n\left(\sqrt{x+\frac1n}- \sqrt{x}\right)$. Show that $f_n$ does not converge uniformly on $(0,\infty)$.

My question:

To showing that whether a sequence of functions is converge uniformly or not, does the first thing to do is by showing that $f_n$ (indeed) converge pointwise to a function, namely, $f:=f(x)=\frac{1}{2\sqrt{x}}$? I have done with this by intuitively and by the usual limit. But, how to show it rigorously? Here's what I tried for the absolute value substraction: $$\begin{align*} |f_n(x)-f(x)| &= \left|\frac{1}{\sqrt{x+\frac1n} + \sqrt{x}} - \frac{1}{2\sqrt{x}}\right| \\ &= \left|\frac{\sqrt{x+\frac1n}-\sqrt{x}}{2\sqrt{x^2 + \frac{x}{n}} + 2x} \right| \\ &\le \left|\frac{\sqrt{x+\frac1n}-\sqrt{x}}{2\sqrt{x^2 + \frac{x}{n}}}\right |. \end{align*}$$ I got stucked here and do not be able yet to found $n \in \Bbb N$ such that the above less than arbitrary $\varepsilon > 0$. Any ideas? Thanks in advanced.

Answer 1

Yes, you must first show that $(f_n)$ (indeed) converge pointwise to a function, say, $f$ with $f:=f(x)=\frac{1}{2\sqrt{x}}$ on $(0,\infty)$, before determining whether $(f_n)$ is converge uniformly or not. To do this rigorously, let $\varepsilon > 0$ and $x \in (0,\infty)$ be arbitrary given. By the Archimedean Principle, we know that there exists $n_0 \in \Bbb N$ such that $n_0 > \frac{1}{4\varepsilon x\sqrt{x}}$. So, for any $n \in \Bbb N$ with $n \ge n_0$, we have \begin{align*} |f_n(x)-f(x)| &= \left|n\left(\sqrt{x+\frac1n} - \sqrt{x} \right) - \frac{1}{2\sqrt{x}} \right| \\ &= \left|\frac{1}{\sqrt{x+\frac1n} + \sqrt{x}} - \frac{1}{2\sqrt{x}}\right| \\ &= \left|\frac{2\sqrt{x} - \sqrt{x+\frac1n} - \sqrt{x}}{2\sqrt{x}\left(\sqrt{x+\frac1n} + \sqrt{x} \right)} \right| \\ &= \left|\frac{\sqrt{x} - \sqrt{x+\frac1n}}{2\sqrt{x}\left(\sqrt{x+\frac1n} + \sqrt{x} \right)} \right| \\ &= \left|\frac{\sqrt{x} - \sqrt{x+\frac1n}}{2\sqrt{x}\left(\sqrt{x+\frac1n} + \sqrt{x} \right)} \right| \cdot \left|\frac{\sqrt{x} + \sqrt{x+\frac1n}}{\sqrt{x} + \sqrt{x+\frac1n}} \right| \\ &= \frac{1}{2n\sqrt{x}\left(\sqrt{x+\frac1n} + \sqrt{x} \right)^2} \\ &\le \frac{1}{2n\sqrt{x}\left(\sqrt{x} + \sqrt{x} \right)^2} \\ &= \frac{1}{2n\sqrt{x}(4x)} \\ &= \frac{1}{8nx\sqrt{x}} \\ &\le \frac{1}{8nx\sqrt{x}} \\ &\le \frac{1}{8n_0x\sqrt{x}} \\ &< \varepsilon. \end{align*} Thus, $(f_n) \to f$ on $(0,\infty)$.

Answer 2

Let $r(x)=\sqrt x$ ($x>0$). Let $x\in(0,\infty)$. It is known that $r'(x)=\frac1{2\sqrt x}$. In other words,$$(\forall\varepsilon>0)(\exists\delta>0):|y-x|<\delta\implies\left|\frac{\sqrt y-\sqrt x}{y-x}-\frac1{2\sqrt x}\right|<\varepsilon.$$Now, take $\varepsilon>0$. Take $\delta>0$ such that$$|y-x|<\delta\implies\left|\frac{\sqrt y-\sqrt x}{y-x}-\frac1{2\sqrt x}\right|<\varepsilon$$Take $N\in\Bbb N$ such that $\frac1N<\delta$. Then\begin{align}n\geqslant N\implies&\frac1n<\delta\\\implies&\left|\frac{\sqrt{x+\frac1n}-\sqrt x}{\frac1n}-\frac1{2\sqrt x}\right|<\varepsilon\\\iff&\left|f_n(x)-\frac1{2\sqrt x}\right|<\varepsilon.\end{align}But the convergence is not uniform. In fact, $\frac1{2\sqrt x}$ is unbounded on $(0,\infty)$, and a sequence of bounded functions ($(\forall n\in\Bbb N):\max|f_n|=\sqrt n$) never converges uniformly to an unbounded function.