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?
Vector Space or Not?
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0@andybenji I looked "one section down, second paragraph", and then what? In fact, extrema do not require derivatives. – 2012-12-07
3 Answers
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
- $(f+g)'(0) = f'(0) + g'(0) = 0 + 0 = 0$.
- The derivative of the zero function is zero, and hence evaluates to 0 at 0.
- $(rf)'(0) = r(f'(0))=r0 = 0$.
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.
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1*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](http://en.wikipedia.org/wiki/Wiener_process#Qualitative_properties). – 2012-12-07
Consider the functions $f(x)=\cases{x&\text{if }x<0\\0&\text{otherwise}}$ and $g(x)=\cases{0&\text{if }x<0\\x&\text{otherwise.}}$
For example, $x^2 + x^3$ and $x^2$ both have relative minima at $0$, but $(x^2 + x^3) - x^2$ does not.
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0A very nice example.:) Many Thanks – 2012-12-02