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 &gt;<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
<br /> \begin{align*}<br /> f(x) = \left\{<br /> \begin{array}{ll}<br /> 0 &amp; \text{ if } x = 0 \\<br /> \dfrac{\pi}{2} &amp; \text{ if } x &gt; 0<br /> \end{array}<br /> \right.<br /> \end{align*}<br />

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 &gt;<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
<br /> \begin{align*}<br /> f(x) = \left\{<br /> \begin{array}{ll}<br /> 0 &amp; \text{ if } x = 0 \\<br /> \dfrac{\pi}{2} &amp; \text{ if } x &gt; 0<br /> \end{array}<br /> \right.<br /> \end{align*}<br />

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

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