Pointwise convergence and uniform convergence

[complexnumber]

Homework Statement

Define the sequence $$\displaystyle f_n : [0,\infty) \to<br /> \left[0,\frac{\pi}{2}\right)$$ by $$f_n(x) := \tan^{-1}(nx), x \geq<br /> 0$$.

Homework Equations

Prove that $$f_n$$ converges pointwise, but not uniformly on
$$[0,\infty)$$.

Prove that $$f_n$$ converges uniformly on $$[t, \infty)$$ for $$t ><br /> 0$$.

The Attempt at a Solution

$$\displaystyle \lim_{n \to \infty} f_n(x) = \lim_{n \to \infty}<br /> \tan^{-1} (nx) = 0$$ for $$x = 0$$.

$$\displaystyle \lim_{n \to \infty} f_n(x) = \lim_{n \to \infty}<br /> \tan^{-1} (nx) = \frac{\pi}{2}$$ for $$x \in (0, \infty)$$.

Let $$\displaystyle f : [0, \infty) \to \left[0, \frac{\pi}{2}<br /> \right)$$ be defined by
$$\begin{align*}<br /> f(x) = \left\{<br /> \begin{array}{ll}<br /> 0 & \text{ if } x = 0 \\<br /> \dfrac{\pi}{2} & \text{ if } x > 0<br /> \end{array}<br /> \right.<br /> \end{align*}$$

Therefore $$f_n$$ converges pointwise to $$f$$.

Is function $$f$$ correct? How can I prove the rest?

 

[HallsofIvy]

Homework Statement

Define the sequence $$\displaystyle f_n : [0,\infty) \to<br /> \left[0,\frac{\pi}{2}\right)$$ by $$f_n(x) := \tan^{-1}(nx), x \geq<br /> 0$$.

Homework Equations

Prove that $$f_n$$ converges pointwise, but not uniformly on
$$[0,\infty)$$.

Prove that $$f_n$$ converges uniformly on $$[t, \infty)$$ for $$t ><br /> 0$$.

The Attempt at a Solution

$$\displaystyle \lim_{n \to \infty} f_n(x) = \lim_{n \to \infty}<br /> \tan^{-1} (nx) = 0$$ for $$x = 0$$.

$$\displaystyle \lim_{n \to \infty} f_n(x) = \lim_{n \to \infty}<br /> \tan^{-1} (nx) = \frac{\pi}{2}$$ for $$x \in (0, \infty)$$.

Let $$\displaystyle f : [0, \infty) \to \left[0, \frac{\pi}{2}<br /> \right)$$ be defined by
$$\begin{align*}<br /> f(x) = \left\{<br /> \begin{array}{ll}<br /> 0 & \text{ if } x = 0 \\<br /> \dfrac{\pi}{2} & \text{ if } x > 0<br /> \end{array}<br /> \right.<br /> \end{align*}$$

Therefore $$f_n$$ converges pointwise to $$f$$.

Is function $$f$$ correct? How can I prove the rest?

Yes, that is correct. Now, just observe that each function $tan^{-1}(nx)$ is continuous for all x while f(x) is not continuous at x= 0. If convergence were uniform, the limit function would also be continuous.

 

[complexnumber]

Thanks. I found a theorem that says if $$f_n$$ converges uniformly to $$f$$ then $$f$$ must be continuous. Is it enough to say that since $$f$$ is
discontinuous at $$0$$, $$f_n$$ cannot converge uniformly to $$f$$ on $$[0,<br /> \infty)$$?

Is the following correct for the second question?

For any $$\varepsilon > 0$$ there must be an $$N \in \mathbb{N}$$ such
that for all $$n > N$$ and for all $$x \in [t, \infty)$$
$$\begin{align*}<br /> \abs{\tan^{-1}(nx) - \frac{\pi}{2}} <& \varepsilon \\<br /> \frac{\pi}{2} - \tan^{-1}(nx) < & \varepsilon \\<br /> \frac{\pi}{2} - \varepsilon < & \tan^{-1}(nx) \\<br /> \tan \left(\frac{\pi}{2} - \varepsilon \right) < & nx \\<br /> \frac{\tan (\frac{\pi}{2} - \varepsilon)}{x} < & n<br /> \end{align*}$$

Therefore choose $$\displaystyle N = \frac{\tan (\frac{\pi}{2} -<br /> \varepsilon)}{t}$$. Then for any $$x \in [t, \infty)$$,
$$\begin{align*}<br /> \abs{f_n(x) - f(x)} = \abs{\tan^{-1}(nx) - \frac{\pi}{2}} \leq<br /> \varepsilon<br /> \end{align*}$$
whenever $$n > N$$. Hence $$f_n$$ converges uniformly to $$f$$ in $$[t,<br /> \infty)$$.

 

[VeeEight]

Thanks. I found a theorem that says if $$f_n$$ converges uniformly to $$f$$ then $$f$$ must be continuous. Is it enough to say that since $$f$$ is
discontinuous at $$0$$, $$f_n$$ cannot converge uniformly to $$f$$ on $$[0,<br /> \infty)$$?

Yes, this is sometimes called the Uniform Limit Theorem and is fine to use (The above poster mentioned it).