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Math Help - Image of bounded operator is closed

  1. #1
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    Image of bounded operator is closed

    Let X, Y be Banach spaces, U \in B(X,Y) be given such that \exists M \in [0, \infty) such that \forall y \in \text{Im}(U), \exists x \in X such that U(x) = y and \|x\|_X \leq M \|y\|_Y. Show that \text{Im}(U) is closed in Y.

    My idea was to use the quotient Banach space of \frac{X}{\text{Ker}(U)} and use the First Isomorphism Theorem, but this isn't very direct. Is there a more direct method for proving this? Note that U isn't necessarily injective.
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  2. #2
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    Quote Originally Posted by Giraffro View Post
    Let X, Y be Banach spaces, U \in B(X,Y) be given such that \exists M \in [0, \infty) such that \forall y \in \text{Im}(U), \exists x \in X such that U(x) = y and \|x\|_X \leq M \|y\|_Y. Show that \text{Im}(U) is closed in Y.

    My idea was to use the quotient Banach space of \frac{X}{\text{Ker}(U)} and use the First Isomorphism Theorem, but this isn't very direct. Is there a more direct method for proving this? Note that U isn't necessarily injective.
    Take a sequence y_n in the image, then let  U(x_n)=y_n. Then

    ||x_n-x_m|| \leq M ||U(x_n-x_m)||=||y_n-y_m|| \rightarrow 0

    So that x_n is Cauchy and hence converges in X to some x. Now then by the continuity of U  y= \lim y_n= \lim U(x_n)= U(\lim x_n)=U(x) and hence y is in the image.
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  3. #3
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    Quote Originally Posted by Focus View Post
    Take a sequence y_n in the image, then let  U(x_n)=y_n. Then

    ||x_n-x_m|| \leq M ||U(x_n-x_m)||=||y_n-y_m|| \rightarrow 0

    So that x_n is Cauchy and hence converges in X to some x. Now then by the continuity of U  y= \lim y_n= \lim U(x_n)= U(\lim x_n)=U(x) and hence y is in the image.
    That's wrong. You don't know that \forall m,n \in \mathbb{N}, \|x_m - x_n\|_X \leq M \|U(x_m-x_n)\|_Y. You can get that \forall m, n \in \mathbb{N}, \exists x_{m,n} \in X such that U(x_{m,n}) = y_m - y_n and \|x_{m,n}\|_X \leq M \|y_m - y_n\|_Y, but that's all.

    x_{m,n} and x_m - x_n can differ by anything in \text{Ker}(U), which is why I used the quotient norm, but there has to be a more direct method.
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  4. #4
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    Quote Originally Posted by Giraffro View Post
    That's wrong. You don't know that \forall m,n \in \mathbb{N}, \|x_m - x_n\|_X \leq M \|U(x_m-x_n)\|_Y. You can get that \forall m, n \in \mathbb{N}, \exists x_{m,n} \in X such that U(x_{m,n}) = y_m - y_n and \|x_{m,n}\|_X \leq M \|y_m - y_n\|_Y, but that's all.

    x_{m,n} and x_m - x_n can differ by anything in \text{Ker}(U), which is why I used the quotient norm, but there has to be a more direct method.
    I think that quotienting by ker(U) is probably essential, because it seems to be the only way to get round the difficulty described above. If you then consider the induced map \hat{U}:X/\ker(U)\to Y, it is injective and still satisfies the condition \|\hat{x}\|_{X/\ker(U)}\leqslant M\|\hat{U}(\hat{x})\|_Y, and so the argument suggested by Focus becomes valid.
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