For each subset of C1 given below, either prove that it is a subspace or show why it is not a subspace.

(a) { f(x) belongs to C1 | f'(pi) = 0 }

(b) { f(x) belongs to C1 | f'(0) = pi }

Question

For each subset of C1 given below, either prove that it is a subspace or show why it is not a subspace.

 

(a) { f(x) belongs to C1 | f'(pi) = 0 }

 

(b) { f(x) belongs to C1 | f'(0) = pi }

anonymous · Answer

Please help

anonymous · Answer

you need to check vector space axioms.
in particular, the second one is not closed under vector addition

anonymous · Answer

I want to proof each one and solution steps please

anonymous · Answer

oh wait it says the derivative right?  ok i still think it is not closed under addition

anonymous · Answer

Yes

anonymous · Answer

ok to prove something is not a vector space, you need only to show that one example violates one axiom

anonymous · Answer

ok I need steps for the proof
Can you give me steps? , please

anonymous · Answer

so for example if 
$$f(x)=x^2+\pi x, g(x)=2x^2+\pi x$$ then 
$$f'(x)=2x+\pi, g'(x)=4x+\pi$$  and so 
$$f'(0)=g'(0)=\pi$$ but 
$$(f+g)'(0)=2\pi\neq \pi$$

anonymous · Answer

if you choose as a function y=constant,it's lie in C1 but not pass throught 0.

anonymous · Answer

satelite73 --> is it part b?

anonymous · Answer

yes that is my example for part b, as to why that is not a vector space, since it is not closed under addition

anonymous · Answer

"0" must be in subsapace

anonymous · Answer

ok I see Thanks, do you have an example for the first one?

anonymous · Answer

what raphaelF said is quicker

anonymous · Answer

i believe the first one is. 
you can check the axioms
zero is certainly in it, and it is also closed. 
most of the other properties are inherited from the definintion of addition of funcitons and multiplying a function by a constant

anonymous · Answer

Can you explain a little bit more

anonymous · Answer

??

anonymous · Answer

do you know the axioms you must check?

anonymous · Answer

addition -  multiply - zeros

anonymous · Answer

I think there are 10

anonymous · Answer

here is a succinct list http://tutorial.math.lamar.edu/Classes/LinAlg/VectorSpaces.aspx

anonymous · Answer

for example $f,g$ have the property that 
$f'(\pi)=g'(\pi)=0$ then since $(f+g)'=f'+g'$ if follows that $(f+g)'(\pi)=0$ this shows it is closed

anonymous · Answer

i think the first is not too:
$$f(x)=1\rightarrow f'(x)=0$$ ,but f(x) not pass throught zero

anonymous · Answer

ok thank you

anonymous · Answer

So both of them not a subspaces riight?

anonymous · Answer

i think i will respectfully disagree.  or at least disagree that because a constant function is in the set that that doesn't make is a vector space. that would disqualify most spaces of functions

anonymous · Answer

I don't understand that

anonymous · Answer

the function does not need to take on the values 0 to be in a vector space. it is rather that the space of functions must include the zero function, which this one does

anonymous · Answer

check the axioms one by one to see if your first set is a vector space. i think that you will find that it is
that is the point of this exercise, to learn how to check the axioms

anonymous · Answer

What do you mean "it's closed"?

anonymous · Answer

if $f\in V$ and $g\in V$ then $f+g\in V$

anonymous · Answer

So the first one is subspace and the second one is not?

anonymous · Answer

that is what i think, yes.  but you should still check the axioms

anonymous · Answer

some you don't really have much to check, like the associative law

anonymous · Answer

ok thank you

anonymous · Answer

lets check one, and lets call our vector space V
lets check that for a scalar c, cf is in V
since $(cf)'=cf'$ it follows that if$f\in V$ i.e. if $f'(\pi)=0$ then likewise $(cf)'(\pi)=0$ and so V is closed under scalar multiplication

anonymous · Answer

yes right