# Bounded linear operator $A$ s.t. $Ax=y$ has a least square solution for each $y$ iff the range of $A$ is closed [closed]

Let $$A: H_1\to H_2$$ be a bounded linear operator, where $$H_1,H_2$$ are Hilbert spaces. Prove that $$Ax=y$$ has a least squares solution for each $$y\in H_2$$ if and only if the range of $$A$$ is closed.

I am having trouble with this problem. I am failing to see why there would not be a least squares solution if the range of $$A$$ is not closed. Nonetheless, I would appreciate any help with formulating a proof.

## closed as off-topic by user21820, Xander Henderson, José Carlos Santos, TheSimpliFire, YuiTo ChengMay 11 at 13:36

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Suppose $$Range \ A$$ is closed. Then consider the closed convex subset $$y- \ Range \ A$$. We know in a Hilbert space any closed convex subset has an element of minimum norm. Thus $$\exists x\in H_1$$ such that $$||y-Ax||$$ is minimized.
For the other part assume $$Range \ A$$ is not closed. Then $$\exists y \in H_2$$ such that $$y$$ is a limit point of $$Range \ A$$ but $$y \notin Range \ A$$. Since $$y$$ is a limit point we have $$d(y,Range \ A):=inf\ \{||y-Ax|| \ ; x\in H_1 \}=0$$ . By existence of least square solution we have $$\hat x \in H_1$$ such that $$||y-A\hat x||=inf\ \{||y-Ax|| \ ; x\in H_1 \}=0$$ But this shows $$y\in Range \ A$$ which contradicts my initial assumption.