Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a question. Suppose that $V$ is a set of all real valued functions that attain its relative maximum or relative minimum at $x=0$. Is V a vector space under the usual operations of addition and scalar multiplications? My guess is it is not a vector space, but I can't able to give a counterexample?

share|cite|improve this question
When talking about real valued functions, it's difficult to use the terms "relative minimum" and "relative maximum". For these to make sense, you're going to need the curves to be at least $C^1$. – andybenji Dec 2 '12 at 8:32
You have not accepted any answer yet! – Mhenni Benghorbal Dec 2 '12 at 8:34
@andybenji: What definition of local minimum/maximum are you using? – wj32 Dec 2 '12 at 8:35
@wj32 I'm using the typical one from calculus: A local maximum is where the derivative goes from positive to negative and a minimum is the opposite. See my answer below for more details. – andybenji Dec 2 '12 at 8:46
@andybenji: I've never seen that (rather strange) definition before. See – wj32 Dec 2 '12 at 9:01
up vote 0 down vote accepted

As I hinted in my comment above, the terms local maximum and local minimum only really make sense when talking about differentiable functions. So here I show that the set of functions with a critical point (not necessarily a local max/min) at 0 (really at any arbitrary point $a \in \mathbb{R}$) is a vector subspace.

If you broaden this slightly to say that $V \subset C^1(-\infty,\infty)$ is the set of differentiable functions on the reals that have a critical point at 0 (i.e. $\forall$ $f \in V$ $f'(0) = 0$).

Then it's simple to show that this is a vector space.

If $f,g \in V$ ($f'(0)=g'(0)=0$), and $r \in \mathbb{R}$ then, and hence

  1. $(f+g)'(0) = f'(0) + g'(0) = 0 + 0 = 0$.
  2. The derivative of the zero function is zero, and hence evaluates to 0 at 0.
  3. $(rf)'(0) = r(f'(0))=r0 = 0$.

And together these imply that $V$ is a vector subspace of $C^1(-\infty,\infty)$.

Robert Israel's answer above is a nice example of why we must define our vector space to have a critical point, not just a max/min at 0.

share|cite|improve this answer
Thank you very much.:) – juniven Dec 2 '12 at 10:15
the terms local maximum and local minimum only really make sense when talking about differentiable functions... Where did you see that? This is far from being true. – Did Dec 7 '12 at 18:33

Consider the functions $$f(x)=\cases{x&\text{if }x<0\\0&\text{otherwise}}$$ and $$g(x)=\cases{0&\text{if }x<0\\x&\text{otherwise.}}$$

share|cite|improve this answer

For example, $x^2 + x^3$ and $x^2$ both have relative minima at $0$, but $(x^2 + x^3) - x^2$ does not.

share|cite|improve this answer
A very nice example.:) Many Thanks – juniven Dec 2 '12 at 10:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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