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(algebra) Proving subspaces functions 
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#1
Oct1110, 06:21 PM

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1. The problem statement, all variables and given/known data
Is U={f E F([tex]\lefta,b\right[/tex]) f(a)=f(b)} a subspace of F([tex]\left a,b \right[/tex]) where F([tex]\left a,b \right[/tex]) is the vector space of real valued functions defined on the interval [a,b]? 2. Relevant equations I know in order for something to be a subspace there are three conditions:  existence of the zero vector [tex]\Theta[/tex]  closed under addition  closed under scalar multiplication 3. The attempt at a solution I tried to determine some boundaries for the function (not 1 to 1, all values will be positive, a and b can be any real number). I'm not really sure how to approach it ... i thought maybe if f(a)=0 then f(b) will also =0. But i don't really know how to prove that or the other two conditions. I would really appreciate someone explaining the steps of how to solve this. Thanks!! 


#2
Oct1110, 06:38 PM

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Now, let f and g be any arbitrary functions in U, and c any scalar. What can you say about (f + g)(a) and (f + g)(b)? What can you say about cf(a) and cf(b)? 


#3
Oct1110, 06:50 PM

P: 4

Oh I understand we can just name two functions since U represents the set of functions...
well (f+g)(a) would = (f+g)(b) and would be equivalent to f(a)+g(a)=f(b)+g(b)? i'm not really sure if i'm right about that. then c*f(a)=c*f(b).. I understand the point of whats happening but not the logic behind whats actually happening, if that makes any sense... 


#4
Oct1110, 06:59 PM

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(algebra) Proving subspaces functions
Can you give reasons for each of the = signs above? What can you conclude about the function f + g? Can you give reasons for each = above? What can you conclude about the function cf? 


#5
Oct1110, 07:12 PM

P: 4

Ok well...
(f+g)(a)=f(a)+g(a) i have no idea i want to say distributive property? f(a)+g(a)= f(b)+g(b) because of the limitations in the question right? if they both have to come to the same solution.. f(a)=f(b) then g(a)=g(b). For some reason all i can think about is extreme value theorem which has nothing to do with this. The function f+g is... no idea. For scalar multiplication it makes sense that if you multiply any scalar by the function its the same as multiplying it inside the function c*f(a)=(cf)(a), and again i'm not really sure but think it has something to do with the f(a)=f(b)... Does the absolute value change anything in solving or does it just imply the functions xvalues stay above the axis? Thank you for your patience! 


#6
Oct1110, 07:30 PM

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P: 21,397

Given two functions f and g, you can create four new functions: f + g, f  g, f*g, and f/g. Besides these four using arithmetic operations, there is also the composition f o g, where (f o g)(x) = f(g(x)). And there's the other order, g o f. 


#7
Oct1110, 07:40 PM

P: 4

Wow I was really not understanding but now really am!!
Oh and i was confused at the notation i thought the square brackets were abs value bars but they're brackets Thank you so much for your help! 


#8
Oct1110, 07:45 PM

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P: 21,397




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