I am self-learning Real Analysis from the text Understanding Analysis, by Stephen Abbott. I would like for someone to

(1) Verify my proof for part (a) of this exercise problem. (2) Do you have any clues for part (b) without giving away the entire solution/proof?

[Abbott 6.3.2] Consider the sequence of functions :

\begin{equation*} h_{n}( x) =\sqrt{x^{2} +\frac{1}{n}} \end{equation*}

(a) Compute the pointwise limit of $\displaystyle ( h_{n})$ and then prove that the convergence is uniform on $\displaystyle \mathbf{R}$.


Fix $\displaystyle x\in \mathbf{R}$. We know that, if $\displaystyle \lim a_{n} =a$, then $\displaystyle \lim \sqrt{a_{n}} =\sqrt{\lim a_{n}} =\sqrt{a}$. Thus:

\begin{equation*} \begin{array}{ c l } \lim _{n\rightarrow \infty } h_{n}( x) & =\lim _{n\rightarrow \infty }\sqrt{x^{2} +\frac{1}{n}}\\ & =\sqrt{\lim _{n\rightarrow \infty }\left( x^{2} +\frac{1}{n}\right)}\\ & =\left[\lim _{n\rightarrow \infty } x^{2} +\lim _{n\rightarrow \infty }\frac{1}{n}\right]^{( 1/2)}\\ & =\sqrt{x^{2}}\\ & =|x| \end{array} \end{equation*}

Consider the expression:

\begin{align*} |h_{n}( x) -h_{m}( x) | & =\left| \sqrt{x^{2} +\frac{1}{n}} -\sqrt{x^{2} +\frac{1}{m}}\right| & \\ & =\frac{\left| \left( x^{2} +\frac{1}{n}\right) -\left( x^{2} +\frac{1}{m}\right)\right| }{\left| \sqrt{x^{2} +\frac{1}{n}} +\sqrt{x^{2} +\frac{1}{m}}\right| } & \\ & =\frac{\left| \frac{1}{n} -\frac{1}{m}\right| }{\sqrt{x^{2} +\frac{1}{n}} +\sqrt{x^{2} +\frac{1}{m}}} \\ & \leq \frac{\left| \frac{1}{n} -\frac{1}{m}\right| }{\frac{1}{\sqrt{n}} +\frac{1}{\sqrt{m}}} & \left\{\because x^{2} \geq 0\right\}\\ & =\frac{\left| \frac{1}{\sqrt{n}} -\frac{1}{\sqrt{m}}\right| \left(\frac{1}{\sqrt{n}} +\frac{1}{\sqrt{m}}\right)}{\left(\frac{1}{\sqrt{n}} +\frac{1}{\sqrt{m}}\right)} & \\ & =\left| \frac{1}{\sqrt{n}} -\frac{1}{\sqrt{m}}\right| & \end{align*}

Pick an arbitrary $\displaystyle \epsilon >0$. Since $\displaystyle \frac{1}{\sqrt{n}}\rightarrow 0$, and convergent sequences are Cauchy, there exists $\displaystyle N( \epsilon ) >0$, such that for all $\displaystyle n >m\geq N$,

\begin{equation*} \left| \frac{1}{\sqrt{n}} -\frac{1}{\sqrt{m}}\right| < \epsilon \end{equation*}

Consequently, by Cauchy criterion for uniform convergence of a sequence of functions, $\displaystyle ( h_{n})$ converges uniformly on $\displaystyle \mathbf{R}$ to $\displaystyle h$.

(b) Note that each $\displaystyle h_{n}$ is differentiable. Show that $\displaystyle g( x) =\lim h_{n} '( x)$ exists for all $\displaystyle x$ and explain how we can be certain that the convergence is not uniform on any neighbourhood of zero.


By Chain rule of differentiation, we have:

\begin{equation*} h_{n} '( x) =\frac{x}{\sqrt{x^{2} +\frac{1}{n}}} \end{equation*} Moreover,

\begin{equation*} \lim h_{n} '( x) =\lim _{n\rightarrow \infty }\frac{x}{\sqrt{x^{2} +\frac{1}{n}}} =\frac{x}{|x|} \end{equation*}


1 Answer 1


Hint: Uniform Convergence preserves certain properties from the sequence of functions to the limit function.

  • $\begingroup$ Can I argue as follows: We are given that the sequence of functions $(h_{n})$ converge pointwise to $\displaystyle h$ and are differentiable on any neighbourhood of zero. By the Differentiable Limit Theorem, if $\displaystyle ( h_{n} ')$ converges uniformly on any neighbourhood of zero to $\displaystyle g$, then $\displaystyle g=h'$. By the contrapositive of the Differentiable Limit Theorem, since $h'$ is not defined at $\displaystyle x=0$, $\displaystyle ( h_{n} ')$ does \ NOT converge uniformly to $\displaystyle h$ on any neighbourhood containing zero. $\endgroup$
    – Quasar
    Oct 30, 2022 at 10:37
  • $\begingroup$ Yes, you could use that or even the fact that uniform convergence preserves continuity $\endgroup$ Oct 30, 2022 at 10:58

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