Exercise 6.3.5

Question

Let
  $$
  g_{n}(x)=\frac{n x+x^{2}}{2 n}
  $$
  and set $g(x)=\lim g_{n}(x)$. Show that $g$ is differentiable in two ways:
  \begin{enumerate}[label=(\alph*)] 
  \item Compute $g(x)$ by algebraically taking the limit as $n \rightarrow \infty$ and then find $g^{\prime}(x)$.
  \item Compute $g_{n}^{\prime}(x)$ for each $n \in \mathbf{N}$ and show that the sequence of derivatives $\left(g_{n}^{\prime}\right)$ converges uniformly on every interval $[-M, M]$. Use Theorem 6.3.3 to conclude $g^{\prime}(x)=\lim g_{n}^{\prime}(x)$.
  \item Repeat parts (a) and (b) for the sequence $f_{n}(x)=\left(n x^{2}+1\right) /(2 n+x)$.

\end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
    \item By inspection $g(x) = x/2$ and $g'(x) = 1/2$.
    \item $g'_n(x) = 1/2 + \frac{x}{n}$ which approaches $\frac{1}{2}$ as $n \to \infty$. Now $\left|g'_n - \frac{1}{2}\right| = \left|\frac{x}{n}\right|$ is bounded by $\frac{M}{n}$ which goes to 0 and is not dependent on $x$, and therefore $(g'_n)$ converges uniformly over $[-M, M]$.
    \item $f(x) = x^2 / 2$, and $f'(x) = x$. We have
        $$f'_n(x) = \frac{4n^2x + nx^2 - 1}{4n^2 + 4nx + x^2} $$
        which approaches $x$ as $n \to \infty$. With some algebra we have
        $$\left|f'_n(x) - x\right| = \left| \frac{x^3 - 3nx^2 - 1}{4n^2 + 4nx + x^2} \right| \leq \frac{M^3 + 3nM^2 + 1}{4n^2 - 4Mn}$$
        which approaches 0 as $n \to \infty$ independent of $x$, and therefore $(f'_n)$ converges uniformly over $[-M, M]$.
 \end{enumerate}

Exercise 6.3.5

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Exercise 6.3.5

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