## 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 2012 at 20:16

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.