I am self-learning Real Analysis from the text, Understanding Analysis by Stephen Abbott. I am having some troubles proving part c) of the below exercise problem. Any hints/suggestions without giving away the entire solution/proof would be extremely helpful.
[Abbott 6.3.5] Let
\begin{equation*}
g_{n}( x) =\frac{nx+x^{2}}{2n}
\end{equation*}
and set $\displaystyle g( x) =\lim g_{n}( x)$. Show that $\displaystyle g$ is differentiable in two ways:(a) Compute $\displaystyle g( x)$ by algebraically taking the limit as $\displaystyle n\rightarrow \infty $ and then find $\displaystyle g'( x)$.
Proof.
Fix $\displaystyle x\in \mathbf{R}$. We have:
\begin{equation*}
\begin{array}{ c l }
\lim _{n\rightarrow \infty } g_{n}( x) & =\lim _{n\rightarrow \infty }\left(\frac{nx+x^{2}}{2n}\right)\\
& =\lim _{n\rightarrow \infty }\left(\frac{x+x^{2} /n}{2}\right)\\
& =\frac{\lim _{n\rightarrow \infty } x+\lim _{n\rightarrow \infty }\frac{x^{2}}{n}}{\lim _{n\rightarrow \infty } 2}\\
& =\frac{x}{2}
\end{array}
\end{equation*}
Thus, the limit function $\displaystyle g( x) =\frac{x}{2}$. The derivative of the limit function is:
\begin{equation*}
g'( x) =\frac{1}{2}
\end{equation*}
(b) Compute $\displaystyle g_{n} '( x)$ for each $\displaystyle n\in \mathbf{N}$ for each $\displaystyle n\in \mathbf{N}$ and show that the sequence of derivatives $\displaystyle ( g_{n} ')$ converges uniformly on every interval $\displaystyle [ -M,M]$. Use the theorem 6.3.3 to conclude that $\displaystyle g'( x) =\lim g_{n} '( x)$.
Proof.
Fix $\displaystyle n\in \mathbf{N}$. By the familiar rules of differentiation:
\begin{equation*}
g_{n} '( x) =\frac{1}{2n}( n+2x) =\frac{1}{2} +\frac{x}{2n}
\end{equation*}
We are interested to prove that $\displaystyle g_{n} '( x)$ converges uniformly on any bounded interval $\displaystyle [ -M,M]$ to the constant function $\displaystyle h( x) =\frac{1}{2}$.
Pick an arbitrary $\displaystyle \epsilon >0$. We are interested to make the distance $\displaystyle |g_{n} '( x) -h( x) |$ smaller than $\displaystyle \epsilon $.
We have:
\begin{equation*}
\begin{array}{ c c }
|g_{n} '( x) -h( x) | & =\left| \frac{x}{2n}\right| \\
& \leq \frac{M}{2n}
\end{array}
\end{equation*}
If we pick $\displaystyle N >\frac{M}{2\epsilon }$, then for all $\displaystyle n\geq N$, and for all $\displaystyle x\in [ -M,M]$, $\displaystyle |g_{n} '( x) -g'( x) |< \epsilon $.
Consequently, the sequence of the derivatives $\displaystyle ( g_{n} ')$ converges uniformly on $\displaystyle [ -M,M]$ to the constant function $\displaystyle h( x) =\lim g_{n} '( x) =\frac{1}{2}$.
We find that the sequence of functions $\displaystyle ( g_{n})$ converge pointwise on the closed interval $\displaystyle [ -M,M]$ to $\displaystyle g$ and are differentiable. Since $\displaystyle ( g_{n} ')$ converges uniformly on $\displaystyle [ -M,M]$ to $\displaystyle h$, by the Differentiable Limit Theorem, it follows that $\displaystyle \lim g_{n} '=h=g'$ on $\displaystyle [ -M,M]$.
(c) Repeat parts (a) and (b) for the sequence $\displaystyle f_{n}( x) =\left( nx^{2} +1\right) /( 2n+x)$.
Proof.
Pointwise convergence of $\displaystyle f_{n}$:
Fix $\displaystyle x\in \mathbf{R}$. We have:
\begin{equation*}
\begin{array}{ c l }
\lim _{n\rightarrow \infty } f_{n}( x) & =\lim _{n\rightarrow \infty }\frac{nx^{2} +1}{2n+x}\\
& =\lim _{n\rightarrow \infty }\frac{\left( x^{2} +1/n\right)}{( 2+x/n)}\\
& =\frac{x^{2}}{2}
\end{array}
\end{equation*}
When I take the derivative $f_n'(x)$ and compute the expression $|f_n'(x)-f'(x)|$, I don't get something very tractable to find an upper bound. I would like to find an upper bound on
$|f_n'(x)-f'(x)|$.
Best Answer
You have$$f_n'(x)-x=-\frac{3 n x^2+x^3+1}{(2 n+x)^2}.$$Take $M\in(0,\infty)$. Then, if $x\in[-M,M]$, you have$$|3nx^2+x^3+1|\leqslant3nM^2+M^3+1$$and, if $n>\frac M2$,$$(2n+x)^2\geqslant(2n-M)^2.$$Can you take it from here?