Let $A\in B(H)$ be a closed range positive operator, and $A^\frac{1}{2}$ its square root. Clearly image of $A$ is a subset of image of $A^\frac{1}{2}~~$($R(A)\subset R(A^\frac{1}{2})$). How do we prove $R(A)= R(A^\frac{1}{2})$? I edite the question
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Is $A$ compact? – Sep 25 '17 at 13:05
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@Kevin, No. Just a positive self adjoint operator – niki Sep 25 '17 at 13:08
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Ok, is $H$ a Hilbert space? – Sep 25 '17 at 13:19
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@Kevin Yes, $H$ is a Hilbert space, and $B(H)$ means the space of bounded linear operators. – niki Sep 25 '17 at 13:21
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@Kevin how can one define "self-adjoint operator" in a non-Hilbert space? – Surb Sep 25 '17 at 13:57
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@surb Just wanted clarification as it was not defined in OP – Sep 25 '17 at 14:05
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@Kevin too bad :), I would have loved to see such a definition. Would be interesting – Surb Sep 25 '17 at 17:21
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@Surb I know of a definition of self adjointedness on a real Banach space if this is interesting to you? – Sep 26 '17 at 07:47
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@Kevin yes, I'd be very interested by that. Does it goes along the lines of : https://mathoverflow.net/questions/154474/self-adjointness-for-banach-spaces ? – Surb Sep 26 '17 at 08:26
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@surb Yes, the Wójcik, paper came to mind! – Sep 26 '17 at 08:40
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This is false. Set $H = \ell^2(\mathbb{N})$ and let $A$ be the multiplication operator defined by
$$ A \left( (x_n)_{n=1}^{\infty} \right) = \left( \frac{x_n}{n} \right)_{n=1}^{\infty}. $$
Then
$$ A^{\frac{1}{2}} \left( (x_n)_{n=1}^{\infty} \right) = \left( \frac{x_n}{\sqrt{n}} \right)_{n=1}^{\infty} $$
and $\left( \frac{1}{n^{1.5}} \right)_{n=1}^{\infty} \in R(A^{\frac{1}{2}})$ but not in $R(A)$.
If the range of $A$ is closed then the result is true. To see why, note that if $x \in \ker(A^{\frac{1}{2}})$ then $Ax = A^{\frac{1}{2}} A^{\frac{1}{2}} x = 0$ so $x \in \ker(A)$ while if $x \in \ker(A)$ then $$ \left< A^{\frac{1}{2}}x, A^{\frac{1}{2}}x \right> = \left< x, Ax \right> = 0 $$
so $x \in \ker(A^{\frac{1}{2}})$. Now,
$$ R(A) \subseteq R \left(A^{\frac{1}{2}} \right) \subseteq \overline{ R \left( A^{\frac{1}{2}} \right)} = \ker \left( A^{\frac{1}{2}} \right)^{\perp} = \ker(A)^{\perp} = \overline{R(A)} = R(A) $$
and so we get $R(A) = R \left(A^{\frac{1}{2}} \right)$.
levap
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BTW, in general you can often find counterexamples easily by playing with multiplication operators. This is not really restrictive as the spectral theorem tells you any self adjoint operator is unitarily equivalent to a multiplication operator. – levap Sep 25 '17 at 14:02
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