# Image of bounded operator is closed

• Jun 3rd 2010, 05:33 AM
Giraffro
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.
• Jun 3rd 2010, 08:20 AM
Focus
Quote:

Originally Posted by Giraffro
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.
• Jun 3rd 2010, 10:25 AM
Giraffro
Quote:

Originally Posted by Focus
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. (Wondering)
• Jun 3rd 2010, 01:03 PM
Opalg
Quote:

Originally Posted by Giraffro
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. (Wondering)

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.