Mixing
A “multi-range” version of Banach-Steinhaus Theorem for Fréchet Spaces
$begingroup$
A general form of the infamous Banach-Steinhaus theorem for Banach spaces can be stated as follows:
Let $X$ be a Banach space, ${Y_alpha }_{alpha in Lambda}$ Be a collection of normed spaces. $T_alpha : X rightarrow Y_alpha $ be a collection of bounded (=continues) linear operators. Then if This collection is pointwise bounded, it is uniformly bounded, in the sense that there exists $M$ such that $forall alpha ; Vert T_alpha Vert < M$
Now a version which is still true for Fréchet spaces (or even barreled spaces) goes as follows: Let $X$ be a Fréchet space, Y be some topological vector space, $T_alpha : X rightarrow Y $ be a collection of continues linear operators. If this collection is pointwise bounded (in the TVS sense) then it is uniformly bounded. In the sense that for all $0 in W subset Y$ there exists $0 in V subset X$ such that $T_alpha (V) subset W$ for all $alpha$.
The main difference (to me) in the two theorems is that in the first one we also had the degree of freedom to choose different $Y_alpha$, and not just $Y$. I wonder whether the following statement is true:
Let $X$ be a Fréchet space, take ${Y_alpha }$ to be some collection of Frechet spaces, $T_alpha : X rightarrow Y_alpha $ a pointwise bounded collection of continues operators. Then for every choice of $r>0$ there exists $delta > 0$ such that $T_alpha(B_delta (0)) subset B_r(0)$ for all $alpha$. (of course I am referring to balls)
I am guessing it's true, but notice I took $Y_alpha$ to be Fréchet spaces, I am not sure how to generalize this to other TVSs.
functional-analysis banach-spaces
edited Jan 17 at 19:56
Bernard
121k740116
asked Jan 17 at 19:37
pitariverpitariver
454113
$endgroup$
add a comment |
$begingroup$
A general form of the infamous Banach-Steinhaus theorem for Banach spaces can be stated as follows:
Let $X$ be a Banach space, ${Y_alpha }_{alpha in Lambda}$ Be a collection of normed spaces. $T_alpha : X rightarrow Y_alpha $ be a collection of bounded (=continues) linear operators. Then if This collection is pointwise bounded, it is uniformly bounded, in the sense that there exists $M$ such that $forall alpha ; Vert T_alpha Vert < M$
Now a version which is still true for Fréchet spaces (or even barreled spaces) goes as follows: Let $X$ be a Fréchet space, Y be some topological vector space, $T_alpha : X rightarrow Y $ be a collection of continues linear operators. If this collection is pointwise bounded (in the TVS sense) then it is uniformly bounded. In the sense that for all $0 in W subset Y$ there exists $0 in V subset X$ such that $T_alpha (V) subset W$ for all $alpha$.
The main difference (to me) in the two theorems is that in the first one we also had the degree of freedom to choose different $Y_alpha$, and not just $Y$. I wonder whether the following statement is true:
Let $X$ be a Fréchet space, take ${Y_alpha }$ to be some collection of Frechet spaces, $T_alpha : X rightarrow Y_alpha $ a pointwise bounded collection of continues operators. Then for every choice of $r>0$ there exists $delta > 0$ such that $T_alpha(B_delta (0)) subset B_r(0)$ for all $alpha$. (of course I am referring to balls)
I am guessing it's true, but notice I took $Y_alpha$ to be Fréchet spaces, I am not sure how to generalize this to other TVSs.
functional-analysis banach-spaces
edited Jan 17 at 19:56
Bernard
121k740116
asked Jan 17 at 19:37
pitariverpitariver
454113
$endgroup$
add a comment |
$begingroup$
A general form of the infamous Banach-Steinhaus theorem for Banach spaces can be stated as follows:
Let $X$ be a Banach space, ${Y_alpha }_{alpha in Lambda}$ Be a collection of normed spaces. $T_alpha : X rightarrow Y_alpha $ be a collection of bounded (=continues) linear operators. Then if This collection is pointwise bounded, it is uniformly bounded, in the sense that there exists $M$ such that $forall alpha ; Vert T_alpha Vert < M$
Now a version which is still true for Fréchet spaces (or even barreled spaces) goes as follows: Let $X$ be a Fréchet space, Y be some topological vector space, $T_alpha : X rightarrow Y $ be a collection of continues linear operators. If this collection is pointwise bounded (in the TVS sense) then it is uniformly bounded. In the sense that for all $0 in W subset Y$ there exists $0 in V subset X$ such that $T_alpha (V) subset W$ for all $alpha$.
The main difference (to me) in the two theorems is that in the first one we also had the degree of freedom to choose different $Y_alpha$, and not just $Y$. I wonder whether the following statement is true:
Let $X$ be a Fréchet space, take ${Y_alpha }$ to be some collection of Frechet spaces, $T_alpha : X rightarrow Y_alpha $ a pointwise bounded collection of continues operators. Then for every choice of $r>0$ there exists $delta > 0$ such that $T_alpha(B_delta (0)) subset B_r(0)$ for all $alpha$. (of course I am referring to balls)
I am guessing it's true, but notice I took $Y_alpha$ to be Fréchet spaces, I am not sure how to generalize this to other TVSs.
functional-analysis banach-spaces
edited Jan 17 at 19:56
Bernard
121k740116
asked Jan 17 at 19:37
pitariverpitariver
454113
$endgroup$
A general form of the infamous Banach-Steinhaus theorem for Banach spaces can be stated as follows:
Let $X$ be a Banach space, ${Y_alpha }_{alpha in Lambda}$ Be a collection of normed spaces. $T_alpha : X rightarrow Y_alpha $ be a collection of bounded (=continues) linear operators. Then if This collection is pointwise bounded, it is uniformly bounded, in the sense that there exists $M$ such that $forall alpha ; Vert T_alpha Vert < M$
Now a version which is still true for Fréchet spaces (or even barreled spaces) goes as follows: Let $X$ be a Fréchet space, Y be some topological vector space, $T_alpha : X rightarrow Y $ be a collection of continues linear operators. If this collection is pointwise bounded (in the TVS sense) then it is uniformly bounded. In the sense that for all $0 in W subset Y$ there exists $0 in V subset X$ such that $T_alpha (V) subset W$ for all $alpha$.
The main difference (to me) in the two theorems is that in the first one we also had the degree of freedom to choose different $Y_alpha$, and not just $Y$. I wonder whether the following statement is true:
Let $X$ be a Fréchet space, take ${Y_alpha }$ to be some collection of Frechet spaces, $T_alpha : X rightarrow Y_alpha $ a pointwise bounded collection of continues operators. Then for every choice of $r>0$ there exists $delta > 0$ such that $T_alpha(B_delta (0)) subset B_r(0)$ for all $alpha$. (of course I am referring to balls)
I am guessing it's true, but notice I took $Y_alpha$ to be Fréchet spaces, I am not sure how to generalize this to other TVSs.
functional-analysis banach-spaces
functional-analysis banach-spaces
edited Jan 17 at 19:56
Bernard
121k740116
asked Jan 17 at 19:37
pitariverpitariver
454113
edited Jan 17 at 19:56
Bernard
121k740116
asked Jan 17 at 19:37
pitariverpitariver
454113
edited Jan 17 at 19:56
Bernard
121k740116
edited Jan 17 at 19:56
Bernard
121k740116
edited Jan 17 at 19:56
Bernard
121k740116
121k740116
asked Jan 17 at 19:37
pitariverpitariver
454113
asked Jan 17 at 19:37
pitariverpitariver
454113
asked Jan 17 at 19:37
pitariverpitariver
454113
454113
add a comment |
add a comment |
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