Exercise 6.3.2

Question

Consider the sequence of functions
  $$
  h_{n}(x)=\sqrt{x^{2}+\frac{1}{n}}.
  $$
  \begin{enumerate}[label=(\alph*)] 
  \item Compute the pointwise limit of $\left(h_{n}\right)$ and then prove that the convergence is uniform on $\mathbf{R}$.
  \item Note that each $h_{n}$ is differentiable. Show $g(x)=\lim h_{n}^{\prime}(x)$ exists for all $x$, and explain how we can be certain that the convergence is not uniform on any neighborhood of zero.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
    \item As $n \to \infty$, $h_n(x) \to \sqrt{x^2} = |x|$. Now, recall that $\sqrt{a} + \sqrt{b} \geq \sqrt{a + b}$ (I think this has been proved earlier, but if not, this is easily shown by squaring both sides); alternatively $\sqrt{a + b} - \sqrt{a} \leq \sqrt{b}$, and so
    $$\left|\sqrt{\left(x^2 + \frac{1}{n}\right)} - \sqrt{x^2}\right| \leq \sqrt{\frac{1}{n}} = \frac{1}{\sqrt{n}}$$
    which we can clearly make less than any $\epsilon > 0$.

\item $$h'_n(x) = \frac{x}{\sqrt{x^2 + 1/n}}$$ which converges to
$$g(x) = \begin{cases}
 1 & x > 0 \
 0 & x = 0 \
 -1 & x < 0
\end{cases}$$
    Define $h(x)$ to be the pointwise limit of $(h_n)$. If the convergence to $g$ was uniform, that would imply that $g_n = h'(x)$ (by the Differentiable Limit Theorem). But from part (a) $h$ is not differentiable at 0, therefore the convergence to $g$ cannot be uniform in a neighborhood around 0.
 \end{enumerate}

Exercise 6.3.2

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Exercise 6.3.2

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