Operator sprzężony
: 15 cze 2023, o 02:01
Znalezc operator sprzezony do
\(\displaystyle{ T :\mathcal{l}_1\rightarrow c_0
\\\\
T(x_1, x_2, ...)= (x_1, x_2, ..., x_n, 0,0,...)
}\)
Rozwiazanie (bardzo prosze o przegladniecie)
\(\displaystyle{ T^*:c_0^*\rightarrow (\mathcal{l}_1)^*
\\\\
T^*: \mathcal{l}_1\rightarrow \mathcal{l}_{\infty} \\\\
T^*(\phi)=\phi\circ T}\)
Niech \(\displaystyle{ x=(x_1,x_2, x_3,...)\in \mathcal{l}_1, \;\phi \in c_0^*\cong\mathcal{l}_1 }\)
\(\displaystyle{ T^*(\phi)(x)= \phi(T(x))=\phi(x_1,x_2,...,x_n, 0,0,...)= \sum_{k=1}^n\phi_kx_k\\\\\\
\eta:=(\phi_1, \phi_2,...,\phi_n,0,0,...)\in \mathcal{l}_{\infty}\cong (\mathcal{l}_1)^* \\\\
\eta(x)=\sum_{k=1}^n\phi_kx_k}\)
Czyli
\(\displaystyle{ T^*(\phi)(x)=\eta(x)\\\\
T^*(\phi)=\eta=(\phi_1, \phi_2,...,\phi_n,0,0,...)
}\)
\(\displaystyle{ T :\mathcal{l}_1\rightarrow c_0
\\\\
T(x_1, x_2, ...)= (x_1, x_2, ..., x_n, 0,0,...)
}\)
Rozwiazanie (bardzo prosze o przegladniecie)
\(\displaystyle{ T^*:c_0^*\rightarrow (\mathcal{l}_1)^*
\\\\
T^*: \mathcal{l}_1\rightarrow \mathcal{l}_{\infty} \\\\
T^*(\phi)=\phi\circ T}\)
Niech \(\displaystyle{ x=(x_1,x_2, x_3,...)\in \mathcal{l}_1, \;\phi \in c_0^*\cong\mathcal{l}_1 }\)
\(\displaystyle{ T^*(\phi)(x)= \phi(T(x))=\phi(x_1,x_2,...,x_n, 0,0,...)= \sum_{k=1}^n\phi_kx_k\\\\\\
\eta:=(\phi_1, \phi_2,...,\phi_n,0,0,...)\in \mathcal{l}_{\infty}\cong (\mathcal{l}_1)^* \\\\
\eta(x)=\sum_{k=1}^n\phi_kx_k}\)
Czyli
\(\displaystyle{ T^*(\phi)(x)=\eta(x)\\\\
T^*(\phi)=\eta=(\phi_1, \phi_2,...,\phi_n,0,0,...)
}\)