0
$\begingroup$

Find the sup-norm, $\|f\|_{\sup}$, if

$$f(x)=\begin{cases} 0, &x \in\mathbb{Q}\\ -x^2, &x\not\in\mathbb{Q} \end{cases}$$

As I look at the graph of $-x^2$, I know it's a decreasing function. I know the sup-norm is infinity. I'm not sure why that is though.

$\endgroup$
  • $\begingroup$ This is very difficult to read. Please edit and type using Latex commands. $\endgroup$ – ADA Apr 24 '17 at 1:59
0
$\begingroup$

Since $$f(x) = \begin{cases} 0 & x \in \mathbb{Q} \\ -x^2 & x \notin \mathbb{Q} \end{cases},$$

we have $$|f(x)| = \begin{cases} 0 & x \in \mathbb{Q} \\ x^2 & x \notin \mathbb{Q} \end{cases}.$$

Notice that $x^2$ can get arbitrary large.

hence $$\left\| f\right\|_\infty= \sup\{|f(x)|: x\in \mathbb{R} \}= \infty$$

$\endgroup$
  • $\begingroup$ Thank you so much! Makes much more sense. $\endgroup$ – ErinA Apr 24 '17 at 2:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.