# The operator preseving two disjoint dense operator ranges invariant

Let $\mathcal{H}$ be a separable infinite dimensional Hilbert space. Suppose that $\mathcal{H}_0\subset\mathcal{H}$ which is a dense proper subspace is the range of some bounded linear operator $T$. It is known that there are many bounded linear operator $X$ preserving $\mathcal{H}_0$ invariant.

Now given disjoint dense proper subspace $\mathcal{H}_1$ and $\mathcal{H}_2$ of $\mathcal{H}$. If there are two bounded linear operators $T_1$ and $T_2$ such that $\mathcal{H}_{i}=Ran(T_i)$ and $X \mathcal{H}_i\subset\mathcal{H}_i$, does the bounded operator $X$ need to be the form $\lambda I$ where $\lambda$ is a complex number?

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Would you please write your question in understandable English and indicate why you are interested in the question and on what spaces? – Bill Johnson Mar 6 '12 at 20:16

## 2 Answers

Of course, not. One can do a purely geometric construction, but, being a Fourier analyst, I'm tempted to present something involving the Fourier transform, so here goes. Let $T_1$ be the multiplication by $e^{-|x|}$ on the space side and let $T_2$ be the multiplication by $e^{-1/x^2}$ on the Fourier side. Than the ranges are dense and essentially disjoint (If $f$ is in the intersection, then $\hat f$ is an analytic function in a horizontal strip that has a very deep zero at the origin, so it is $0$ identically). However, any convolution with compactly supported $L^1$ kernel preserves both ranges.

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Dear Professor Fedja, it is a good hint. Thank you! I am an oeperator algebrast and am trying to solve some problems which involve the Fourier analysis and the operator theory. Would you like to give me your email so that I can contact with you?

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To be honest, I prefer a public discussion. Just post your questions here or on AoPS (the SE forum moves too fast and is too overcrowded with people looking for help with their homeworks to run the discussion there, but AoPS College playground is a nice place if you think that your questions are not OK for MO). – fedja Mar 21 '12 at 2:49