Existence of solution of some functional equation involving integral
$begingroup$
I want to prove that there exists always solutions to this equation
$$g(x) = intlimits_a^b {f(s,x)ds} $$
where $g in {L^2}(0,1)$ is given and $f in {L^2}((a,b) times (0,1))$ is the uknown, $0<a<b<1$ are just to positive constants.
If I consider the operator $$eqalign{
& T:{L^2}((a,b) times (0,1)) to {L^2}(0,1) cr
& f to intlimits_a^b {f(s,x)ds} cr} $$
the surjectivity peoblem turn to the existence of a positive constant $c$ such that
$${left| {{T^*}h} right|_{{L^2}(0,1)}} geqslant c{left| h right|_{{L^2}((a,b) times (0,1))}}$$
The formal ajoint of $T$ is given by ${T^*}h =h$, therefore, we can write
$${left| {{T^*}h} right|_{{L^2}(0,1)}} = intlimits_0^1 {{h^2}(s)ds = } intlimits_a^b {intlimits_0^1 {{h^2}(s)dsdx = } } frac{1}{{b - a}}{left| h right|_{{L^2}((a,b) times (0,1))}}$$.
Am I write? thank you.
real-analysis functional-analysis operator-theory functional-equations adjoint-operators
asked Jan 25 at 15:48
GustaveGustave
734211
$endgroup$
add a comment |
$begingroup$
I want to prove that there exists always solutions to this equation
$$g(x) = intlimits_a^b {f(s,x)ds} $$
where $g in {L^2}(0,1)$ is given and $f in {L^2}((a,b) times (0,1))$ is the uknown, $0<a<b<1$ are just to positive constants.
If I consider the operator $$eqalign{
& T:{L^2}((a,b) times (0,1)) to {L^2}(0,1) cr
& f to intlimits_a^b {f(s,x)ds} cr} $$
the surjectivity peoblem turn to the existence of a positive constant $c$ such that
$${left| {{T^*}h} right|_{{L^2}(0,1)}} geqslant c{left| h right|_{{L^2}((a,b) times (0,1))}}$$
The formal ajoint of $T$ is given by ${T^*}h =h$, therefore, we can write
$${left| {{T^*}h} right|_{{L^2}(0,1)}} = intlimits_0^1 {{h^2}(s)ds = } intlimits_a^b {intlimits_0^1 {{h^2}(s)dsdx = } } frac{1}{{b - a}}{left| h right|_{{L^2}((a,b) times (0,1))}}$$.
Am I write? thank you.
real-analysis functional-analysis operator-theory functional-equations adjoint-operators
asked Jan 25 at 15:48
GustaveGustave
734211
$endgroup$
1
$begingroup$
Why not just take $f(s,x) = g(x)/(b-a)$?
$endgroup$
– Robert Israel
Jan 25 at 16:04
$begingroup$
Yes, this is one possibility, but I want to know if my proof is correct.
$endgroup$
– Gustave
Jan 25 at 16:06
add a comment |
$begingroup$
I want to prove that there exists always solutions to this equation
$$g(x) = intlimits_a^b {f(s,x)ds} $$
where $g in {L^2}(0,1)$ is given and $f in {L^2}((a,b) times (0,1))$ is the uknown, $0<a<b<1$ are just to positive constants.
If I consider the operator $$eqalign{
& T:{L^2}((a,b) times (0,1)) to {L^2}(0,1) cr
& f to intlimits_a^b {f(s,x)ds} cr} $$
the surjectivity peoblem turn to the existence of a positive constant $c$ such that
$${left| {{T^*}h} right|_{{L^2}(0,1)}} geqslant c{left| h right|_{{L^2}((a,b) times (0,1))}}$$
The formal ajoint of $T$ is given by ${T^*}h =h$, therefore, we can write
$${left| {{T^*}h} right|_{{L^2}(0,1)}} = intlimits_0^1 {{h^2}(s)ds = } intlimits_a^b {intlimits_0^1 {{h^2}(s)dsdx = } } frac{1}{{b - a}}{left| h right|_{{L^2}((a,b) times (0,1))}}$$.
Am I write? thank you.
real-analysis functional-analysis operator-theory functional-equations adjoint-operators
asked Jan 25 at 15:48
GustaveGustave
734211
$endgroup$
I want to prove that there exists always solutions to this equation
$$g(x) = intlimits_a^b {f(s,x)ds} $$
where $g in {L^2}(0,1)$ is given and $f in {L^2}((a,b) times (0,1))$ is the uknown, $0<a<b<1$ are just to positive constants.
If I consider the operator $$eqalign{
& T:{L^2}((a,b) times (0,1)) to {L^2}(0,1) cr
& f to intlimits_a^b {f(s,x)ds} cr} $$
the surjectivity peoblem turn to the existence of a positive constant $c$ such that
$${left| {{T^*}h} right|_{{L^2}(0,1)}} geqslant c{left| h right|_{{L^2}((a,b) times (0,1))}}$$
The formal ajoint of $T$ is given by ${T^*}h =h$, therefore, we can write
$${left| {{T^*}h} right|_{{L^2}(0,1)}} = intlimits_0^1 {{h^2}(s)ds = } intlimits_a^b {intlimits_0^1 {{h^2}(s)dsdx = } } frac{1}{{b - a}}{left| h right|_{{L^2}((a,b) times (0,1))}}$$.
Am I write? thank you.
real-analysis functional-analysis operator-theory functional-equations adjoint-operators
real-analysis functional-analysis operator-theory functional-equations adjoint-operators
asked Jan 25 at 15:48
GustaveGustave
734211
asked Jan 25 at 15:48
GustaveGustave
734211
asked Jan 25 at 15:48
GustaveGustave
734211
asked Jan 25 at 15:48
GustaveGustave
734211
asked Jan 25 at 15:48
GustaveGustave
734211
734211
1
$begingroup$
Why not just take $f(s,x) = g(x)/(b-a)$?
$endgroup$
– Robert Israel
Jan 25 at 16:04
$begingroup$
Yes, this is one possibility, but I want to know if my proof is correct.
$endgroup$
– Gustave
Jan 25 at 16:06
add a comment |
1
$begingroup$
Why not just take $f(s,x) = g(x)/(b-a)$?
$endgroup$
– Robert Israel
Jan 25 at 16:04
$begingroup$
Yes, this is one possibility, but I want to know if my proof is correct.
$endgroup$
– Gustave
Jan 25 at 16:06
1
1
$begingroup$
Why not just take $f(s,x) = g(x)/(b-a)$?
$endgroup$
– Robert Israel
Jan 25 at 16:04
$begingroup$
Why not just take $f(s,x) = g(x)/(b-a)$?
$endgroup$
– Robert Israel
Jan 25 at 16:04
$begingroup$
Yes, this is one possibility, but I want to know if my proof is correct.
$endgroup$
– Gustave
Jan 25 at 16:06
$begingroup$
Yes, this is one possibility, but I want to know if my proof is correct.
$endgroup$
– Gustave
Jan 25 at 16:06
add a comment |
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$begingroup$
Why not just take $f(s,x) = g(x)/(b-a)$?
$endgroup$
– Robert Israel
Jan 25 at 16:04
$begingroup$
Yes, this is one possibility, but I want to know if my proof is correct.
$endgroup$
– Gustave
Jan 25 at 16:06