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Given the vector space, $ C(-\infty,\infty)$ as the set of all continuous functions that are always continuous, is the set of all exponential functions, $U=\{a^x\mid a \ge 1 \}$, a subspace of the given vector space?

As far as I'm aware, proving a subspace of a given vector space only requires you to prove closure under addition and scalar multiplication, but I'm kind of at a loss as to how to do this with exponential functions (I'm sure it's way simpler than I'm making it).

My argument so far is that the set $U$ is a subset of the set of all differentiable functions, which itself is a subset of $C(-\infty,\infty)$, but I doubt that argument would hold up on my test, given how we've tested for subspaces in class (with closure).

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  • $\begingroup$ If you say exponential functions, did you mean to write $a^x$ instead of $x^a$? $\endgroup$
    – StackTD
    Jul 15, 2017 at 14:55
  • $\begingroup$ I did and I've corrected that, thank you! $\endgroup$
    – wzbillings
    Jul 15, 2017 at 14:56
  • $\begingroup$ The zero in $C(-\infty,\infty)$ is the constant function the always returns zero. That function is not in $U$. $\endgroup$
    – user463383
    Jul 15, 2017 at 14:58
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    $\begingroup$ What is $a^x$ at $x=0$? What is $a^x+b^x$ at $x=0$? Can it equal some $c^x$? $\endgroup$ Jul 15, 2017 at 14:58

3 Answers 3

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As originally written, no, since $0\notin U.$ In addition, your $U$ seems to be the set of monic monomials in $x,$ not exponential functions, which would instead have the form $a^x$ for some positive constant $a.$ Even then, this would not be a subspace, for exactly the same reason.

Edit: Another simple approach would be to show that $U$ (in either version) is not closed under scalar multiplication.

Showing failure to be closed under addition is more difficult in both cases, but not too tricky for the original $U.$

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  • $\begingroup$ Didn't think about it not being closed under scalar multiplication--thanks for that. I wasn't super confident about the logic I was using, thank you for the explanation! $\endgroup$
    – wzbillings
    Jul 15, 2017 at 17:17
  • $\begingroup$ You're very welcome. It's good to keep in mind that some sets may be closed under one, but not the other. For example, the set $\bigl\{(x,y)\in\Bbb R^2:x>0\bigr\}$ is closed under addition, but not scalar multiplication; the set $\bigl\{(x,y)\in\Bbb R^2:x=0\text{ or }y=0\bigr\}$ is closed under scalar multiplication, but not addition. $\endgroup$ Jul 15, 2017 at 17:33
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Not, it is not a subspace. For instance the null function does not belong to $ U$.

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(...) but I doubt that argument would hold up on my test, given how we've tested for subspaces in class (with closure).

Hint: closure under addition means that if you would take two such functions, say $a^x$ and $b^x$, then their sum $a^x+b^x$ should be in the set as well. That function is only in the set if $a^x+b^x = c^x$ for some $c$; is that possible (in general)?

Or the fast and simple test: does the set contain the zero element (in this case: the zero function)?

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