Exercise 6.3.3

Question

Consider the sequence of functions
  $$
  f_{n}(x)=\frac{x}{1+n x^{2}} .
  $$
  \begin{enumerate}[label=(\alph*)] 
  \item Find the points on $\mathbf{R}$ where each $f_{n}(x)$ attains its maximum and minimum value. Use this to prove $\left(f_{n}\right)$ converges uniformly on $\mathbf{R}$. What is the limit function?
  \item Let $f=\lim f_{n}$. Compute $f_{n}^{\prime}(x)$ and find all the values of $x$ for which $f^{\prime}(x)=\lim f_{n}^{\prime}(x) .$
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
    \item $f_n$ is differentiable on $\mathbf{R}$, so by the Interior Extremum Theorem the maximum and minimum values will appear where $f'_n(x) = 0$. We have
    $$f'_n(x) = \frac{1-nx^2}{\left(1 + nx^2\right)^2}$$
    which is zero at $x = \pm 1/\sqrt{n}$. Plugging these values back into $f_n(x)$ we get that $\left|f_n(x)\right| \leq \frac{1}{2\sqrt{n}}$. Clearly this forces $f_n$ to converge uniformly to 0.
    \item $f(x) = f'(x) = 0$. We have
    $$\lim_{n \to \infty} f'_n(x) = \frac{1 - nx^2}{1 + 2nx^2 + n^2 x^4} = \frac{\frac{1}{n} - x^2}{\frac{1}{n} + 2x^2 + nx^4} = 0$$
    and therefore $f'(x) = \lim f'_n(x)$ everywhere.
 \end{enumerate}

Exercise 6.3.3

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Exercise 6.3.3

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