Injectivity of integral operators
0
$begingroup$
Let $K:L^2[0,1]^{d_1}to L^2[0,1]^{d_2}$ be integral operator
$$(Kf)(y) = int f(x)k(x,y)d x.$$
If $d_1>d_2$ is it possible for $K$ to be injective?, e.g. let's take $d_1=2,d_2=1$.
More generally, does the injectivity of $K$ impose any restrictions on $d_1,d_2$.
functional-analysis functional-inequalities
asked Nov 28 '18 at 17:06
LionvilleLionville
334112
$endgroup$
add a comment |
0
$begingroup$
Let $K:L^2[0,1]^{d_1}to L^2[0,1]^{d_2}$ be integral operator
$$(Kf)(y) = int f(x)k(x,y)d x.$$
If $d_1>d_2$ is it possible for $K$ to be injective?, e.g. let's take $d_1=2,d_2=1$.
More generally, does the injectivity of $K$ impose any restrictions on $d_1,d_2$.
functional-analysis functional-inequalities
asked Nov 28 '18 at 17:06
LionvilleLionville
334112
$endgroup$
$begingroup$
Is there any information on $k(x,y)$? For instance, in the trivial case when $k(x,y)$ does not depend on $y$ and vanishes for all $x in (0,0.5)$, then $Kf(y) =0 forall y$ for some non-zero function $f$, so the operator is not injective.
$endgroup$
– Yuxin Wang
Nov 29 '18 at 1:43
$begingroup$
Thank you. I don't have any conditions. I need an example of injective $K$ when $d_1>d_2$ (or at least in knowing whether this is possible). I'm not interested in counterexamples.
$endgroup$
– Lionville
Nov 29 '18 at 1:45
$begingroup$
Well, what are your thoughts on the question so far?
$endgroup$
– jgon
Nov 29 '18 at 2:30
add a comment |
0
0
0
$begingroup$
Let $K:L^2[0,1]^{d_1}to L^2[0,1]^{d_2}$ be integral operator
$$(Kf)(y) = int f(x)k(x,y)d x.$$
If $d_1>d_2$ is it possible for $K$ to be injective?, e.g. let's take $d_1=2,d_2=1$.
More generally, does the injectivity of $K$ impose any restrictions on $d_1,d_2$.
functional-analysis functional-inequalities
asked Nov 28 '18 at 17:06
LionvilleLionville
334112
$endgroup$
Let $K:L^2[0,1]^{d_1}to L^2[0,1]^{d_2}$ be integral operator
$$(Kf)(y) = int f(x)k(x,y)d x.$$
If $d_1>d_2$ is it possible for $K$ to be injective?, e.g. let's take $d_1=2,d_2=1$.
More generally, does the injectivity of $K$ impose any restrictions on $d_1,d_2$.
functional-analysis functional-inequalities
functional-analysis functional-inequalities
asked Nov 28 '18 at 17:06
LionvilleLionville
334112
asked Nov 28 '18 at 17:06
LionvilleLionville
334112
edited Nov 29 '18 at 1:45
Lionville
asked Nov 28 '18 at 17:06
LionvilleLionville
334112
asked Nov 28 '18 at 17:06
LionvilleLionville
334112
asked Nov 28 '18 at 17:06
LionvilleLionville
334112
334112
$begingroup$
Is there any information on $k(x,y)$? For instance, in the trivial case when $k(x,y)$ does not depend on $y$ and vanishes for all $x in (0,0.5)$, then $Kf(y) =0 forall y$ for some non-zero function $f$, so the operator is not injective.
$endgroup$
– Yuxin Wang
Nov 29 '18 at 1:43
$begingroup$
Thank you. I don't have any conditions. I need an example of injective $K$ when $d_1>d_2$ (or at least in knowing whether this is possible). I'm not interested in counterexamples.
$endgroup$
– Lionville
Nov 29 '18 at 1:45
$begingroup$
Well, what are your thoughts on the question so far?
$endgroup$
– jgon
Nov 29 '18 at 2:30
add a comment |
$begingroup$
Is there any information on $k(x,y)$? For instance, in the trivial case when $k(x,y)$ does not depend on $y$ and vanishes for all $x in (0,0.5)$, then $Kf(y) =0 forall y$ for some non-zero function $f$, so the operator is not injective.
$endgroup$
– Yuxin Wang
Nov 29 '18 at 1:43
$begingroup$
Thank you. I don't have any conditions. I need an example of injective $K$ when $d_1>d_2$ (or at least in knowing whether this is possible). I'm not interested in counterexamples.
$endgroup$
– Lionville
Nov 29 '18 at 1:45
$begingroup$
Well, what are your thoughts on the question so far?
$endgroup$
– jgon
Nov 29 '18 at 2:30
$begingroup$
Is there any information on $k(x,y)$? For instance, in the trivial case when $k(x,y)$ does not depend on $y$ and vanishes for all $x in (0,0.5)$, then $Kf(y) =0 forall y$ for some non-zero function $f$, so the operator is not injective.
$endgroup$
– Yuxin Wang
Nov 29 '18 at 1:43
$begingroup$
Is there any information on $k(x,y)$? For instance, in the trivial case when $k(x,y)$ does not depend on $y$ and vanishes for all $x in (0,0.5)$, then $Kf(y) =0 forall y$ for some non-zero function $f$, so the operator is not injective.
$endgroup$
– Yuxin Wang
Nov 29 '18 at 1:43
$begingroup$
Thank you. I don't have any conditions. I need an example of injective $K$ when $d_1>d_2$ (or at least in knowing whether this is possible). I'm not interested in counterexamples.
$endgroup$
– Lionville
Nov 29 '18 at 1:45
$begingroup$
Thank you. I don't have any conditions. I need an example of injective $K$ when $d_1>d_2$ (or at least in knowing whether this is possible). I'm not interested in counterexamples.
$endgroup$
– Lionville
Nov 29 '18 at 1:45
$begingroup$
Well, what are your thoughts on the question so far?
$endgroup$
– jgon
Nov 29 '18 at 2:30
$begingroup$
Well, what are your thoughts on the question so far?
$endgroup$
– jgon
Nov 29 '18 at 2:30
add a comment |
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$begingroup$
Is there any information on $k(x,y)$? For instance, in the trivial case when $k(x,y)$ does not depend on $y$ and vanishes for all $x in (0,0.5)$, then $Kf(y) =0 forall y$ for some non-zero function $f$, so the operator is not injective.
$endgroup$
– Yuxin Wang
Nov 29 '18 at 1:43
$begingroup$
Thank you. I don't have any conditions. I need an example of injective $K$ when $d_1>d_2$ (or at least in knowing whether this is possible). I'm not interested in counterexamples.
$endgroup$
– Lionville
Nov 29 '18 at 1:45
$begingroup$
Well, what are your thoughts on the question so far?
$endgroup$
– jgon
Nov 29 '18 at 2:30