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Determine whether the set of functions is a vector space

  1. Feb 5, 2013 #1
    1. The problem statement, all variables and given/known data

    Let I = [a,b], a closed interval. With addition and scalar multiplication as defined for all real-valued continuous functions defined on I, determine which of the following sets of functions is a vector space.

    a) All continuous functions, f, such that f(a)=f(b)

    b) All continuous functions, f, such that f(a)=f(b)=0

    c) All continuous functions, f, such that f(a)=f(b)=-1

    d) All continuous functions, f, such that f((a+b)/2)=0

    e) All constant functions on [a,b]


    2. Relevant equations

    The operation + (vector addition) must satisfy the following conditions:

    Closure: If u and v are any vectors in V, then the sum u + v belongs to V.
    (1) Commutative law: For all vectors u and v in V, u + v = v + u
    (2) Associative law: For all vectors u, v, w in V, u + (v + w) = (u + v) + w
    (3) Additive identity: The set V contains an additive identity element, denoted by 0, such that for any vector v in V, 0 + v = v and v + 0 = v.
    (4) Additive inverses: For each vector v in V, the equations v + x = 0 and x + v = 0 have a solution x in V, called an additive inverse of v, and denoted by - v.
    The operation · (scalar multiplication) is defined between real numbers (or scalars) and vectors, and must satisfy the following conditions:
    Closure: If v in any vector in V, and c is any real number, then the product c · v belongs to V.
    (5) Distributive law: For all real numbers c and all vectors u, v in V, c · (u + v) = c · u + c · v
    (6) Distributive law: For all real numbers c, d and all vectors v in V, (c+d) · v = c · v + d · v
    (7) Associative law: For all real numbers c,d and all vectors v in V, c · (d · v) = (cd) · v
    (8) Unitary law: For all vectors v in V, 1 · v = v

    3. The attempt at a solution

    All I can think of is showing whether or not each function satisfies all those properties. The properties were not given but this is what I found on the internet. Our teacher told us we should have learned this back in calculus however I placed out of the classes at my school due to AP's and I've never seen this. Its not in my book and we don't have notes because it was suppose to "jog our memory". If someone could just show me how to determine if one of those problems is a vector field so I can figure out the rest that would be great!
     
  2. jcsd
  3. Feb 5, 2013 #2

    Dick

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    You don't have to worry about most of those properties. They are known to be true for numbers or functions. Just concentrate on the closure properties. You want to show that if f(x) and g(x) are in the set and a is a scalar then f(x)+g(x) is in the set and af(x) is in the set. Start with the first one. If f(a)=f(b) and g(a)=g(b) is it true that f(a)+g(a)=f(b)+g(b)?
     
  4. Feb 5, 2013 #3
    Okay so for a) if f(a)=f(b) and g(a)=g(b) then f(a)+g(a)=f(b)+g(b) so a is a vector space?
    Then it'd be the same for b) since f(a)=f(b)=0 then g(a)=g(b)=0 so f(a)+g(a)=f(b)+g(b)=0 and therefore b is a vector space also?

    then for c it shouldn't be a vector space since f(a)=f(b)=-1 and if g(a)=g(b)=-1 then f(a)+g(a) =f(b)+g(b)=-2 which isnt -1 so its not a vector space?
     
  5. Feb 5, 2013 #4

    Dick

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    You need to check closure under the scalar product as well, but yes, you are on the right track. And you should probably note that if f is continuous and g is continuous then f+g and af are also continuous. a) and b) are vector spaces. c) is not. Continue.
     
  6. Feb 6, 2013 #5

    HallsofIvy

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    To clarify what Dick said, to be a vector space, it must satisfy both closure of the sum and closure of the scalar product. For (a) and (b) you have shown the first but not yet the second. For (c), since it does not have closure of the sum of vectors, it doesn't matter whether the scalar product is closed and you do NOT have to look at that.
     
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