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Consider the vector space $C([-\pi,\pi])$ of all continuous functions on the interval $[-\pi,\pi]$ equipped with the norm $$||f||= \max_{x\in[-\pi,\pi]}\{|f(x)|\}.$$

sketch the closed ball of center $\sin x$ and radius one.

$\textbf{Attempt:}$ I'm not sure what this question really is asking. Wouldn't the closed ball simply be two the maxima of $|\sin x|$ in the interval $[-\pi,\pi]$. Meaning just the point 1?

user119264
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    The ball is a subset of $C([-\pi,\pi])$. It is a set of functions, not numbers. Once you understand the elements of the ball, it will make more sense what it means to "sketch" it. – Andrew Dudzik May 16 '15 at 01:20
  • Are you using the definition of the closed ball centered at a point to conclude this? (I.e., look at the definition, and try to apply it). – pjs36 May 16 '15 at 01:21
  • @pjs36 I'm not sure what you mean. This was an exam question, I took today, so can you elaborate what you mean? – user119264 May 16 '15 at 01:24
  • I meant that it doesn't seem like you're using the definition of a ball. I usually see the "Ball centered at $z$, of radius $r$", defined as $B_r(z) = {x : \lVert x - z \rVert = r }$. So the tricky part is figuring out how that definition (or the definition you're using, however the distance is written) applies to your situation. – pjs36 May 16 '15 at 01:41

3 Answers3

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Joining to other answers, this is a "sketch".

$B(\sin x, 1)$

One could consider continuous functions as their graphs. Then the ball is a set of functions which graphs in this banded pipe (as @copper.hat has noted). In case of closed ball graphs could touch boundary of the pipe.

Nikita Evseev
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The closed ball $\bar{B}(\sin,1)$ consists of continuous $f$ such that $|f(x)-\sin x| \le 1$ for all $x$.

To visualise, form the curves $x \mapsto \sin x \pm 1$, then the closed ball is all the continuous functions whose graph lies inside the 'envelope' formed by these curves.

Or, imagine putting a pipe of radius one around the sine curve, then all continuous $f$ that lie in or on the pipe form the closed ball around $\sin$.

copper.hat
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The basic idea is that you need to resort to the definition of a closed ball.

In a metric space with distance function $d(x, y)$, the closed ball centered at $x_0$ with radius $r$ is given by

$$\overline{B_r}(x_0) = \{x : d(x, x_0) \leq r\}$$

(note that in my comment to the question, I had mistakenly used "$=$", when I meant "$\leq$"; it's all about the definitions!). In a normed space with norm $\lVert \cdot \rVert$, it's typical to define the distance function

$$d(x, y) \overset{\text{def}}{=} \lVert x - y \rVert,$$

so that the definition of closed ball above becomes $$\overline{B_r}(x_0) = \{x : \lVert x - x_0 \rVert \leq r\}.$$

Your job is to rewrite this, using your norm on the function space $C([-\pi, \pi])$ for your particular example. Then you can start to think about what functions are in your closed ball.

As far as sketching goes, it's not very easy to sketch anything besides functions in any function space. In this case, you can sketch the region of $\Bbb R^2$ in which the graphs of functions in $\overline{B_1}\big(\sin(x)\big)$ have to live.

pjs36
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