Compact Integral Operators induced by positive Kernels
Let $K$ be a compact operator induced by the kernel $k(s,t)in L^2([0,1])^2$ with $k(s,t)>0$. Prove that $|K|<1$ if and only if $(I-K)$ has a bounded inverse $(I-K)^{-1}$ which is induced by a positive kernel.
I am mainly confused how $(I-K)^{-1}$ is induced by a kernel at all. If we assume $|K|<1$, then $(I-K)^{-1}=sum_{j=0}^infty K^j$ is indeed bounded, but why is it induced by a kernel? I could maybe show that $sum_{j=1}^infty K^j$ is induced by a positive kernel, but is $I$ itself even an integral operator, i.e. induced by a kernel? Am I misunderstanding something? I would be really happy if someone could explain to me how this is meant.
compact-operators integral-equations
asked 2 days ago
Analysis801Analysis801
1175
add a comment |
Let $K$ be a compact operator induced by the kernel $k(s,t)in L^2([0,1])^2$ with $k(s,t)>0$. Prove that $|K|<1$ if and only if $(I-K)$ has a bounded inverse $(I-K)^{-1}$ which is induced by a positive kernel.
I am mainly confused how $(I-K)^{-1}$ is induced by a kernel at all. If we assume $|K|<1$, then $(I-K)^{-1}=sum_{j=0}^infty K^j$ is indeed bounded, but why is it induced by a kernel? I could maybe show that $sum_{j=1}^infty K^j$ is induced by a positive kernel, but is $I$ itself even an integral operator, i.e. induced by a kernel? Am I misunderstanding something? I would be really happy if someone could explain to me how this is meant.
compact-operators integral-equations
asked 2 days ago
Analysis801Analysis801
1175
add a comment |
Let $K$ be a compact operator induced by the kernel $k(s,t)in L^2([0,1])^2$ with $k(s,t)>0$. Prove that $|K|<1$ if and only if $(I-K)$ has a bounded inverse $(I-K)^{-1}$ which is induced by a positive kernel.
I am mainly confused how $(I-K)^{-1}$ is induced by a kernel at all. If we assume $|K|<1$, then $(I-K)^{-1}=sum_{j=0}^infty K^j$ is indeed bounded, but why is it induced by a kernel? I could maybe show that $sum_{j=1}^infty K^j$ is induced by a positive kernel, but is $I$ itself even an integral operator, i.e. induced by a kernel? Am I misunderstanding something? I would be really happy if someone could explain to me how this is meant.
compact-operators integral-equations
asked 2 days ago
Analysis801Analysis801
1175
Let $K$ be a compact operator induced by the kernel $k(s,t)in L^2([0,1])^2$ with $k(s,t)>0$. Prove that $|K|<1$ if and only if $(I-K)$ has a bounded inverse $(I-K)^{-1}$ which is induced by a positive kernel.
I am mainly confused how $(I-K)^{-1}$ is induced by a kernel at all. If we assume $|K|<1$, then $(I-K)^{-1}=sum_{j=0}^infty K^j$ is indeed bounded, but why is it induced by a kernel? I could maybe show that $sum_{j=1}^infty K^j$ is induced by a positive kernel, but is $I$ itself even an integral operator, i.e. induced by a kernel? Am I misunderstanding something? I would be really happy if someone could explain to me how this is meant.
compact-operators integral-equations
compact-operators integral-equations
asked 2 days ago
Analysis801Analysis801
1175
asked 2 days ago
Analysis801Analysis801
1175
asked 2 days ago
Analysis801Analysis801
1175
asked 2 days ago
Analysis801Analysis801
1175
asked 2 days ago
Analysis801Analysis801
1175
1175
add a comment |
add a comment |
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