Adjoint of a surjective bounded linear operator on a Hilbert Space
$begingroup$
Let $H$ be a Hilbert Space and let $T in BL(H, K)$ be surjective then prove that the adjoint of $T$ is bounded below.
Since $T$ is surjective, so for any $y in K, exists$ $x in H$ such that $Tx = y$ $implies ||y||=||Tx|| leq ||T||.||x|| implies ||y|| leq c||x||$, for some $c>0$.
We also define a unique $T^*: K to H$ such that $langle Tx, yrangle = xT^*y$, as the adjoint of $T$.
It would be very helpful if I could get an insight of how to proceed from here. Thanks.
functional-analysis hilbert-spaces
asked Nov 28 '18 at 16:46
JackTJackT
319111
$endgroup$
add a comment |
$begingroup$
Let $H$ be a Hilbert Space and let $T in BL(H, K)$ be surjective then prove that the adjoint of $T$ is bounded below.
Since $T$ is surjective, so for any $y in K, exists$ $x in H$ such that $Tx = y$ $implies ||y||=||Tx|| leq ||T||.||x|| implies ||y|| leq c||x||$, for some $c>0$.
We also define a unique $T^*: K to H$ such that $langle Tx, yrangle = xT^*y$, as the adjoint of $T$.
It would be very helpful if I could get an insight of how to proceed from here. Thanks.
functional-analysis hilbert-spaces
asked Nov 28 '18 at 16:46
JackTJackT
319111
$endgroup$
add a comment |
$begingroup$
Let $H$ be a Hilbert Space and let $T in BL(H, K)$ be surjective then prove that the adjoint of $T$ is bounded below.
Since $T$ is surjective, so for any $y in K, exists$ $x in H$ such that $Tx = y$ $implies ||y||=||Tx|| leq ||T||.||x|| implies ||y|| leq c||x||$, for some $c>0$.
We also define a unique $T^*: K to H$ such that $langle Tx, yrangle = xT^*y$, as the adjoint of $T$.
It would be very helpful if I could get an insight of how to proceed from here. Thanks.
functional-analysis hilbert-spaces
asked Nov 28 '18 at 16:46
JackTJackT
319111
$endgroup$
Let $H$ be a Hilbert Space and let $T in BL(H, K)$ be surjective then prove that the adjoint of $T$ is bounded below.
Since $T$ is surjective, so for any $y in K, exists$ $x in H$ such that $Tx = y$ $implies ||y||=||Tx|| leq ||T||.||x|| implies ||y|| leq c||x||$, for some $c>0$.
We also define a unique $T^*: K to H$ such that $langle Tx, yrangle = xT^*y$, as the adjoint of $T$.
It would be very helpful if I could get an insight of how to proceed from here. Thanks.
functional-analysis hilbert-spaces
functional-analysis hilbert-spaces
asked Nov 28 '18 at 16:46
JackTJackT
319111
asked Nov 28 '18 at 16:46
JackTJackT
319111
asked Nov 28 '18 at 16:46
JackTJackT
319111
asked Nov 28 '18 at 16:46
JackTJackT
319111
asked Nov 28 '18 at 16:46
JackTJackT
319111
319111
add a comment |
add a comment |
1 Answer
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$begingroup$
Suppose that $T^*$ is not bounded below. There exists a sequence $(f_nin H^*) $ such that $|f_n|=1$ and $|T^*(f_n) |<1/n$. This implies that
$sup_{|x|=1}|T(f_n(x))|<1/n$.
Since $T$ is surjective open mapping theorem implies that $T(B_K(0,1))$ contains $B_H(0,s)$. Since $|f_n|=1$, There exists $y_nin H$ such that $|y_n|=1$ and $|f_n(y_n)|geq 1/2$. Since $T$ is surjective, $sy_n=T(x_n) $. We can suppose that $x_nin B_K(0,1)$. This implies that $|f_n(T(x_n)|=|T^*(f_n) (x_n)|geq s/2$. We deduce that $|T^*(f_n) |geq s/2$ for every $n$. Contradiction.
answered Nov 28 '18 at 17:49
Tsemo AristideTsemo Aristide
57.6k11444
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1 Answer
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$begingroup$
Suppose that $T^*$ is not bounded below. There exists a sequence $(f_nin H^*) $ such that $|f_n|=1$ and $|T^*(f_n) |<1/n$. This implies that
$sup_{|x|=1}|T(f_n(x))|<1/n$.
Since $T$ is surjective open mapping theorem implies that $T(B_K(0,1))$ contains $B_H(0,s)$. Since $|f_n|=1$, There exists $y_nin H$ such that $|y_n|=1$ and $|f_n(y_n)|geq 1/2$. Since $T$ is surjective, $sy_n=T(x_n) $. We can suppose that $x_nin B_K(0,1)$. This implies that $|f_n(T(x_n)|=|T^*(f_n) (x_n)|geq s/2$. We deduce that $|T^*(f_n) |geq s/2$ for every $n$. Contradiction.
answered Nov 28 '18 at 17:49
Tsemo AristideTsemo Aristide
57.6k11444
$endgroup$
add a comment |
$begingroup$
Suppose that $T^*$ is not bounded below. There exists a sequence $(f_nin H^*) $ such that $|f_n|=1$ and $|T^*(f_n) |<1/n$. This implies that
$sup_{|x|=1}|T(f_n(x))|<1/n$.
Since $T$ is surjective open mapping theorem implies that $T(B_K(0,1))$ contains $B_H(0,s)$. Since $|f_n|=1$, There exists $y_nin H$ such that $|y_n|=1$ and $|f_n(y_n)|geq 1/2$. Since $T$ is surjective, $sy_n=T(x_n) $. We can suppose that $x_nin B_K(0,1)$. This implies that $|f_n(T(x_n)|=|T^*(f_n) (x_n)|geq s/2$. We deduce that $|T^*(f_n) |geq s/2$ for every $n$. Contradiction.
answered Nov 28 '18 at 17:49
Tsemo AristideTsemo Aristide
57.6k11444
$endgroup$
add a comment |
$begingroup$
Suppose that $T^*$ is not bounded below. There exists a sequence $(f_nin H^*) $ such that $|f_n|=1$ and $|T^*(f_n) |<1/n$. This implies that
$sup_{|x|=1}|T(f_n(x))|<1/n$.
Since $T$ is surjective open mapping theorem implies that $T(B_K(0,1))$ contains $B_H(0,s)$. Since $|f_n|=1$, There exists $y_nin H$ such that $|y_n|=1$ and $|f_n(y_n)|geq 1/2$. Since $T$ is surjective, $sy_n=T(x_n) $. We can suppose that $x_nin B_K(0,1)$. This implies that $|f_n(T(x_n)|=|T^*(f_n) (x_n)|geq s/2$. We deduce that $|T^*(f_n) |geq s/2$ for every $n$. Contradiction.
answered Nov 28 '18 at 17:49
Tsemo AristideTsemo Aristide
57.6k11444
$endgroup$
Suppose that $T^*$ is not bounded below. There exists a sequence $(f_nin H^*) $ such that $|f_n|=1$ and $|T^*(f_n) |<1/n$. This implies that
$sup_{|x|=1}|T(f_n(x))|<1/n$.
Since $T$ is surjective open mapping theorem implies that $T(B_K(0,1))$ contains $B_H(0,s)$. Since $|f_n|=1$, There exists $y_nin H$ such that $|y_n|=1$ and $|f_n(y_n)|geq 1/2$. Since $T$ is surjective, $sy_n=T(x_n) $. We can suppose that $x_nin B_K(0,1)$. This implies that $|f_n(T(x_n)|=|T^*(f_n) (x_n)|geq s/2$. We deduce that $|T^*(f_n) |geq s/2$ for every $n$. Contradiction.
answered Nov 28 '18 at 17:49
Tsemo AristideTsemo Aristide
57.6k11444
answered Nov 28 '18 at 17:49
Tsemo AristideTsemo Aristide
57.6k11444
answered Nov 28 '18 at 17:49
Tsemo AristideTsemo Aristide
57.6k11444
answered Nov 28 '18 at 17:49
Tsemo AristideTsemo Aristide
57.6k11444
57.6k11444
add a comment |
add a comment |
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