Hi, I have a linear bounded operator A between tow Hilbert spaces with a dense domain and dense range, is it surjective? If the answer is no, are there more conditions on A to be surjective?
Dear Hafdallah,
A can be not surjective. For instance, let A: l_2 -> l_2 be as follows
A(x_1,..., x_n, ...)=(x_1/2,..., x_n/2^n, ...) (x_n is divided by 2^n).
It is clear that the set F of all finite sequences Y=(y_1,...,y_m,0,0,0,...) is contained in Im A (m depends on Y and can be arbitrary). Clearly, F is dense in l_2.
But there is no pre-image for, say, Y=(1,1/2,...,1/n,...) \in l_2.
You should be more specific about A.
Best regards, Yuri
Dear Hafdallah,
A can be not surjective. For instance, let A: l_2 -> l_2 be as follows
A(x_1,..., x_n, ...)=(x_1/2,..., x_n/2^n, ...) (x_n is divided by 2^n).
It is clear that the set F of all finite sequences Y=(y_1,...,y_m,0,0,0,...) is contained in Im A (m depends on Y and can be arbitrary). Clearly, F is dense in l_2.
But there is no pre-image for, say, Y=(1,1/2,...,1/n,...) \in l_2.
You should be more specific about A.
Best regards, Yuri
Dear Hafdallah,
The assertion is not true, by various reasons. Here is one of them: following Theorem 2.1, Chapter III, Section 2, page 77 from "Topological Vector Spaces" by H. H. Schaefer (third edition), for a continuous linear map u from a complete metrizable topological vector space L into another such space M, u having dense range, either u(L) is of first (Baire) category, or else u(L)=M and u is a topological homomorphism (it is an "open" map: the image u(D) of any open subset D of L is an open subsets of M). Hence there exists continuous linear mappings with dense range, which are not surjective, and for which the first variant for u(L) holds. I recall that M cannot be of first category. On the contrary, it is of second Baire category, being metrizable and complete.
Best regards, Octav
Dear Hafdallah,
As Yuri's example shows, a linear continuous operator between two Hilbert spaces need not be onto, but may have a dense range.
One may ask when, having a dense range, a linear continuous operator is onto? Here is a possible answer:
Let L be a linear continuous operator between two Banach spaces X and Y. If L(X) is dense in Y and there exists a constant M>0 such that for each y in L(X) there exists an x in X such that M||x||
It is interesting that my answer, which can be verified by anyone, was down voted. For me this fact is a surprise, and it is hard to give it a reasonable explanation. I presume that it is for fun. Otherwise, the person who did it has not enough mathematical knowledge.
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