Exercise 6.3.1

Question

Consider the sequence of functions defined by
  $$
  g_{n}(x)=\frac{x^{n}}{n} .
  $$
  \begin{enumerate}[label=(\alph*)] 
  \item Show $\left(g_{n}\right)$ converges uniformly on $[0,1]$ and find $g=\lim g_{n}$. Show that $g$ is differentiable and compute $g^{\prime}(x)$ for all $x \in[0,1]$.
  \item Now, show that $\left(g_{n}^{\prime}\right)$ converges on $[0,1]$. Is the convergence uniform? Set $h=\lim g_{n}^{\prime}$ and compare $h$ and $g^{\prime}$. Are they the same?
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
    \item I claim that $(g_n) \to 0$. This can be seen by noting that for $x \in [0,1]$, $0 \leq x^n \leq 1$ and so $0 \leq g_n(x) \leq 1/n$. Thus for any $\epsilon > 0$, any $n > N = 1/\epsilon$ will force $|g_n(x) - 0| < \epsilon$. $g(x)=0$ is obviously differentiable, with its derivative just $0$.
    \item $g'(n) = x^{n-1}$. Using a similar argument,
$$h(x) = \lim_{n\to\infty} g'_n(x) = \begin{cases}
    0 & x \in [0,1) \
    1 & x = 1
\end{cases}$$
which is not equal to $g'(x)$ at $1$. Tne convergence is not uniform. For $\epsilon = 0.5$ and any given $n$, choosing $1 > x > \sqrt[N]{\epsilon}$ leads to $g_n(x) > \epsilon$, preventing uniform convergence.
 \end{enumerate}

Exercise 6.3.1

Answers

Comments

Exercise 6.3.1

Answers

Comments

Add answer