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Math Help - Prove of closed subspace

  1. #1
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    Prove of closed subspace

    Given W[0,1]:\mathbb{R} \rightarrow \mathbb{C} defined as the space of continuously differentiable functions with inner product (f,g)_{W}=\int^{1}_{0} f(t)\overline{g(t)}+f'(t)\overline{g'(t)} \ dt.

    Let f \in W[0,1] and show:

    1. (f,\cosh) = f(1)\sinh(1)
    2. M=\{f \in W[0,1]:f(1)=0\} is a closed subspace of W[0,1]

    I am somewhat uncertain of my solution, thus asking here...

    (f,cosh)
    = \int^{1}_{0} f(t) \overline{\cosh(t)}+f'(t)\overline{\cosh'(t)} \ dt
    = [F(t)\cosh(t)]^{1}_{0} - \int^{1}_{0}F(t)\overline{\cosh'(t)} \ dt + [f(t)\overline{\cosh'(t)}]^{1}_{0} - \int^{1}_{0} f(t) \overline{\cosh''(t)} \ dt
    =[f(t)\overline{\cosh'(t)}]^{1}_{0}
    =f(1)\sinh(1).

    Correct?

    Further...
    f(t)=0 for t \in [0,1] belongs to M thus M non-empty.

    Assume f,g \in M and c \in \mathbb{C}. Then (f+g)(1) = f(1)+g(1) = 0. Thus (f+g) \in M and M is closed under addition. Also (f+cg)(1) = f(1)+(cg)(1) = f(1)+c(g)(1) = 0. Thus (f+cg) \in M and M closed under scalar multiplication.

    Correct?

    But how do I show M is a closed subspace? Is it enough to to state it is a continous function defined on a closed and limited interval thus attaining both its minimum and maximum value?
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  2. #2
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    Quote Originally Posted by Cere Kong View Post
    Given W[0,1]:\mathbb{R} \rightarrow \mathbb{C} defined as the space of continuously differentiable functions with inner product (f,g)_{W}=\int^{1}_{0} f(t)\overline{g(t)}+f'(t)\overline{g'(t)} \ dt.

    Let f \in W[0,1] and show:

    1. (f,\cosh) = f(1)\sinh(1)
    2. M=\{f \in W[0,1]:f(1)=0\} is a closed subspace of W[0,1]

    ...

    But how do I show M is a closed subspace? Is it enough to to state it is a continuous function defined on a closed and limited interval thus attaining both its minimum and maximum value?
    Hint: To show that M is closed, use part (1), which tells you that f(1)=0\ \Longleftrightarrow\  (f,\cosh)_W = 0.
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