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I'm having a terrible time understanding subspaces (and, well, linear algebra in general). I'm presented with the problem:

Determine whether the following are subspaces of C[-1,1]:

a) The set of functions f in C[-1,1] such that f(-1)=f(1)

e) The set of functions f in C[-1,1] such that f(-1)=0 or f(1)=0

I'm not sure that I even completely understand the question, let alone how to solve the problem. Before I go about ripping my hair out, can someone perhaps explain to me how to approach this problem?

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  • $\begingroup$ For (e), you can find two counter example functions in that set that, when added together, gives a function not in that set. $\endgroup$ – peterwhy Oct 1 '13 at 22:38
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    $\begingroup$ There are two conditions you need to check. 1) that the zero of $C[-1,1]$ is also an element of the presumed subspace. 2) that given any two functions, $f$ and $g$ in the presumed subspace and any two numbers $a$ and $b$, we have that $af+bg$ is also in the presumed subspace. Checking 1 is usually easy. To check 2 we just have to assume $f$ and $g$ satisfy the condition defining the presumed subspace and show that $af+bg$ also satisfy this condition. $\endgroup$ – OR. Oct 1 '13 at 22:41
  • $\begingroup$ In theory I think that makes sense, but how do I go about performing those operations? As I mentioned, I'm having trouble understand how to implement the theories as well. $\endgroup$ – rphello101 Oct 1 '13 at 22:44
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    $\begingroup$ Dear rphell, You can rephrase @ABC 's condition (2) by breaking it into two pieces: (2a) check that if $f$ and $g$ are in the putative subspace, then so is $f + g$ ; (2b) check that if $f$ is in the putative subspace, and $c$ is a scalar, then $c f$ is in the putative subspace. So, if $f(-1) = f(1)$ and also $g(-1) = g(1)$, does $f+g$ have the same property? What about $cf$? Or, if at least one of $f(-1)$ or $f(1)$ equals $0$, and the same with $g$, what about $f+g$? What about a scalar multiple of $f$? Regards, $\endgroup$ – Matt E Oct 1 '13 at 23:14
  • $\begingroup$ P.S. If you're not sure how to check these, perhaps your question is more basic? Do you know what $f+g$ means? Or $c f$? (Where $f$ and $g$ are functions and $c$ is a scalar, i.e. a number.) If not, then maybe that's what you should be asking about. $\endgroup$ – Matt E Oct 1 '13 at 23:18
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Examples

An example member of $C[-1, 1]$ is $f_1(x) = (x+1)(x-1),\quad -1 \leq x \leq 1$, it's also an example of a function that satisfies $f_1(-1) = f_1(1)$.

$f_2(x) = (x+1)(2-x), \quad -1 \leq x \leq 1$, is an example of a member that has $f_2(-1) = 0$ or $f_2(1) = 0$ because $f_2(-1) = 0$ but $f_2(1) = 2 \neq 0$.

Looking at examples always helps to understand and also can provide counterexamples when you're proving something false. When it's true, you ultimately have to be able to prove it generally, like with (a).

I suppose I should mention, just in case it introduces confusion that $f_1$ also satisfies the (b) condition as it is zero at both extremes. Logical binary "or" only fails if both arguments fail.

$f_3 = x^2 + 2$ satisfies (a) but not (b) as $f_3(-1) = f_3(1) = 3 \neq 0$.

Answer

For (a), if $f(-1) = f(1)$ then $kf(-1) = kf(1)$.

Also, if $g(-1) = g(1)$ then $f(-1) + g(-1) = f(1) + g(1)$.

With problems like this, just grab a piece of paper and a pen and jump in. You find once you start writing things out they'll fall into place. Don't just try and think it out.

For (b) the "or" is the clue, try $f$ being zero only on the left and $g$ being zero only on the right. What happens when you add them together? Problems with function spaces like this usually involve point wise operations so you're really just adding numbers together and seeing what happens.

Addendum 1

Answer to (b): If $f(-1) = 0$, $f(1) = a \neq 0$, $g(-1) = b \neq 0$, $g(1) = 0$, these both satisfy the requirements for at least one of these values to be zero.

\begin{align*} (f+g)(-1) &= f(-1) + g(-1) = 0 + b = b \neq 0 \\ (f+g)(1) &= f(1) + g(1) = a + 0 = a \neq 0 \end{align*}

And so $f+g$ does not satisfy the requirements as both values are non zero. This is a counterexample. We only need to show one where it's not a closed subset, so it's not a subspace.

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  • $\begingroup$ I think I understand how to go about proving the first. I know the second one is not a subspace, but I'm still a little confused on how to show a case where it doesn't hold true. Any chance you can explain it a bit more? $\endgroup$ – rphello101 Oct 3 '13 at 20:44
  • $\begingroup$ I hope the addendum to my answer makes sense. If not, let me know. I will be starting my own mathematics website for publishing various articles pertaining to my teaching of maths in general. I've almost finished an article introducing the idea of function spaces. It's a bit long to paste in here but I'll come back soon and paste a link once it's ready. $\endgroup$ – Geoff Pointer Oct 3 '13 at 23:08
  • $\begingroup$ That answer does make sense. Thanks for adding it. That would actually be a great resource. Any idea when you'll post it so I know when to look? $\endgroup$ – rphello101 Oct 4 '13 at 1:55
  • $\begingroup$ Well, I'm supposed to be an honours student and do course work and a project, but I'm a hopeless sidetrack merchant. But, I enjoy teaching maths at this level and beyond which is why I find time for this site. I'll be starting my maths blog pretty soon and if it goes to plan it should be up within the next week. Cheers. $\endgroup$ – Geoff Pointer Oct 4 '13 at 5:15
  • $\begingroup$ @rphello101 Okay, it's pretty rudimentary because I'm creating my web site from scratch in order to learn and keep in tune with the latest from the global academic community rather than use some fancy software machine which might be generating bad html. I'm sure it's good for me. Try My Maths Blog. My first article there is inspired by this question. My next challenge is to make the PDFs scale better and be more readable. I hope it's useful. Cheers. $\endgroup$ – Geoff Pointer Oct 4 '13 at 7:33

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