Mixing
Discuss the existence and uniqueness of the following operator equation
$begingroup$
The question says
Let $A:X to Y$ where $X,Y$ are normed spaces.
$(i)$ Discuss the uniqueness and existence of a solution of the operator equation $Ax=y$ where $x in X$ and $y in Y$ for different operators $A$
$(ii)$ Discuss for the different values of $lambda$ the existence and uniqueness of this operator equation $Ax=lambda x -y$
I think for $(i)$ the only way for the solution $x$ to exist is that $A$ becomes bijective in order to have an inverse st $x=A^{-1}y$ and in this case the solution is unique. I don't seem to have any other ideas concerning $(i)$.
for $(ii)$ I think we can never discuss it unless $A:X to X$. Moreover, I assumed the following about $A$
if $A in L(X)$ where $L(X)$ is the space of linear bounded operators and $|A| leq lambda$ then we can re-write the equation as $(Ilambda - A)x=-y$
where $x= -frac{1}{lambda} sum_{n=0}^{infty} frac{A^n}{lambda^n}(y)$ this proves the existence and for every choice of $lambda$ we get a new solution.
I don't have any other ideas.
(**Note. I did not study compact operators **)
real-analysis functional-analysis proof-verification proof-writing operator-theory
asked Jan 6 at 12:59
Dreamer123Dreamer123
27629
$endgroup$
add a comment |
$begingroup$
The question says
Let $A:X to Y$ where $X,Y$ are normed spaces.
$(i)$ Discuss the uniqueness and existence of a solution of the operator equation $Ax=y$ where $x in X$ and $y in Y$ for different operators $A$
$(ii)$ Discuss for the different values of $lambda$ the existence and uniqueness of this operator equation $Ax=lambda x -y$
I think for $(i)$ the only way for the solution $x$ to exist is that $A$ becomes bijective in order to have an inverse st $x=A^{-1}y$ and in this case the solution is unique. I don't seem to have any other ideas concerning $(i)$.
for $(ii)$ I think we can never discuss it unless $A:X to X$. Moreover, I assumed the following about $A$
if $A in L(X)$ where $L(X)$ is the space of linear bounded operators and $|A| leq lambda$ then we can re-write the equation as $(Ilambda - A)x=-y$
where $x= -frac{1}{lambda} sum_{n=0}^{infty} frac{A^n}{lambda^n}(y)$ this proves the existence and for every choice of $lambda$ we get a new solution.
I don't have any other ideas.
(**Note. I did not study compact operators **)
real-analysis functional-analysis proof-verification proof-writing operator-theory
asked Jan 6 at 12:59
Dreamer123Dreamer123
27629
$endgroup$
1
$begingroup$
Note that for (i) existence would follow from surjectivity of $A$ and uniqueness from injectivity. Both together follow from bijectivity but it is not necessary for $A$ to have a continuous inverse. For (ii) as you note correctly the expression $lambda x - y$ needs to make sense. So you need to be able to add vectors of $X$ with vectors of $Y$, which means either $Y$ is a subspace of $X$ or $X$ is a subspace of $Y$. At any rate as you note you are essentially asking the question from before for $(lambda - A)x=y$, so apply the results from before here.
$endgroup$
– s.harp
Jan 8 at 13:58
add a comment |
$begingroup$
The question says
Let $A:X to Y$ where $X,Y$ are normed spaces.
$(i)$ Discuss the uniqueness and existence of a solution of the operator equation $Ax=y$ where $x in X$ and $y in Y$ for different operators $A$
$(ii)$ Discuss for the different values of $lambda$ the existence and uniqueness of this operator equation $Ax=lambda x -y$
I think for $(i)$ the only way for the solution $x$ to exist is that $A$ becomes bijective in order to have an inverse st $x=A^{-1}y$ and in this case the solution is unique. I don't seem to have any other ideas concerning $(i)$.
for $(ii)$ I think we can never discuss it unless $A:X to X$. Moreover, I assumed the following about $A$
if $A in L(X)$ where $L(X)$ is the space of linear bounded operators and $|A| leq lambda$ then we can re-write the equation as $(Ilambda - A)x=-y$
where $x= -frac{1}{lambda} sum_{n=0}^{infty} frac{A^n}{lambda^n}(y)$ this proves the existence and for every choice of $lambda$ we get a new solution.
I don't have any other ideas.
(**Note. I did not study compact operators **)
real-analysis functional-analysis proof-verification proof-writing operator-theory
asked Jan 6 at 12:59
Dreamer123Dreamer123
27629
$endgroup$
The question says
Let $A:X to Y$ where $X,Y$ are normed spaces.
$(i)$ Discuss the uniqueness and existence of a solution of the operator equation $Ax=y$ where $x in X$ and $y in Y$ for different operators $A$
$(ii)$ Discuss for the different values of $lambda$ the existence and uniqueness of this operator equation $Ax=lambda x -y$
I think for $(i)$ the only way for the solution $x$ to exist is that $A$ becomes bijective in order to have an inverse st $x=A^{-1}y$ and in this case the solution is unique. I don't seem to have any other ideas concerning $(i)$.
for $(ii)$ I think we can never discuss it unless $A:X to X$. Moreover, I assumed the following about $A$
if $A in L(X)$ where $L(X)$ is the space of linear bounded operators and $|A| leq lambda$ then we can re-write the equation as $(Ilambda - A)x=-y$
where $x= -frac{1}{lambda} sum_{n=0}^{infty} frac{A^n}{lambda^n}(y)$ this proves the existence and for every choice of $lambda$ we get a new solution.
I don't have any other ideas.
(**Note. I did not study compact operators **)
real-analysis functional-analysis proof-verification proof-writing operator-theory
real-analysis functional-analysis proof-verification proof-writing operator-theory
asked Jan 6 at 12:59
Dreamer123Dreamer123
27629
asked Jan 6 at 12:59
Dreamer123Dreamer123
27629
edited Jan 6 at 16:54
Dreamer123
asked Jan 6 at 12:59
Dreamer123Dreamer123
27629
asked Jan 6 at 12:59
Dreamer123Dreamer123
27629
asked Jan 6 at 12:59
Dreamer123Dreamer123
27629
27629
1
$begingroup$
Note that for (i) existence would follow from surjectivity of $A$ and uniqueness from injectivity. Both together follow from bijectivity but it is not necessary for $A$ to have a continuous inverse. For (ii) as you note correctly the expression $lambda x - y$ needs to make sense. So you need to be able to add vectors of $X$ with vectors of $Y$, which means either $Y$ is a subspace of $X$ or $X$ is a subspace of $Y$. At any rate as you note you are essentially asking the question from before for $(lambda - A)x=y$, so apply the results from before here.
$endgroup$
– s.harp
Jan 8 at 13:58
add a comment |
1
$begingroup$
Note that for (i) existence would follow from surjectivity of $A$ and uniqueness from injectivity. Both together follow from bijectivity but it is not necessary for $A$ to have a continuous inverse. For (ii) as you note correctly the expression $lambda x - y$ needs to make sense. So you need to be able to add vectors of $X$ with vectors of $Y$, which means either $Y$ is a subspace of $X$ or $X$ is a subspace of $Y$. At any rate as you note you are essentially asking the question from before for $(lambda - A)x=y$, so apply the results from before here.
$endgroup$
– s.harp
Jan 8 at 13:58
1
1
$begingroup$
Note that for (i) existence would follow from surjectivity of $A$ and uniqueness from injectivity. Both together follow from bijectivity but it is not necessary for $A$ to have a continuous inverse. For (ii) as you note correctly the expression $lambda x - y$ needs to make sense. So you need to be able to add vectors of $X$ with vectors of $Y$, which means either $Y$ is a subspace of $X$ or $X$ is a subspace of $Y$. At any rate as you note you are essentially asking the question from before for $(lambda - A)x=y$, so apply the results from before here.
$endgroup$
– s.harp
Jan 8 at 13:58
$begingroup$
Note that for (i) existence would follow from surjectivity of $A$ and uniqueness from injectivity. Both together follow from bijectivity but it is not necessary for $A$ to have a continuous inverse. For (ii) as you note correctly the expression $lambda x - y$ needs to make sense. So you need to be able to add vectors of $X$ with vectors of $Y$, which means either $Y$ is a subspace of $X$ or $X$ is a subspace of $Y$. At any rate as you note you are essentially asking the question from before for $(lambda - A)x=y$, so apply the results from before here.
$endgroup$
– s.harp
Jan 8 at 13:58
add a comment |
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$begingroup$
Note that for (i) existence would follow from surjectivity of $A$ and uniqueness from injectivity. Both together follow from bijectivity but it is not necessary for $A$ to have a continuous inverse. For (ii) as you note correctly the expression $lambda x - y$ needs to make sense. So you need to be able to add vectors of $X$ with vectors of $Y$, which means either $Y$ is a subspace of $X$ or $X$ is a subspace of $Y$. At any rate as you note you are essentially asking the question from before for $(lambda - A)x=y$, so apply the results from before here.
$endgroup$
– s.harp
Jan 8 at 13:58