How to prove that $(C( F), langle.,.rangle_2)$ is a pre-Hilbert space?
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Let $F$ be a surface. For all continuous functions $f,g in C(F) $ define
$$ langle f,grangle_2 := int_F f(x)g(x), dx $$
I'm struggling to show, that $ (C(F),langle.,.rangle_2) $ is a pre-hilbert space. Can you help me out? :) Any help will be very appreciated.
Showing that $<f,g>$ is a scalarproduct:
1.
$ 0 leq $ $<f,f> $
and if $ <f,f>=0 $
$ leftrightarrow int_F f(x)f(x) do(x)=0 $
then $f(x) =0 $
2.
$<f+h,g>= int_F (f+h)(x)g(x) do(x) = int_F f(x)g(x)+int_f h(x)g(x)= int_F f(x)g(x) do(x) + int h(x)g(x) do(x)= <f,g>+<h,g> $
3.
$ < lambda f,g>= lambda <f,g> $ is clear
real-analysis inner-product-space
asked Dec 5 at 18:36
wondering1123
10011
|
show 1 more comment
up vote
2
down vote
favorite
Let $F$ be a surface. For all continuous functions $f,g in C(F) $ define
$$ langle f,grangle_2 := int_F f(x)g(x), dx $$
I'm struggling to show, that $ (C(F),langle.,.rangle_2) $ is a pre-hilbert space. Can you help me out? :) Any help will be very appreciated.
Showing that $<f,g>$ is a scalarproduct:
1.
$ 0 leq $ $<f,f> $
and if $ <f,f>=0 $
$ leftrightarrow int_F f(x)f(x) do(x)=0 $
then $f(x) =0 $
2.
$<f+h,g>= int_F (f+h)(x)g(x) do(x) = int_F f(x)g(x)+int_f h(x)g(x)= int_F f(x)g(x) do(x) + int h(x)g(x) do(x)= <f,g>+<h,g> $
3.
$ < lambda f,g>= lambda <f,g> $ is clear
real-analysis inner-product-space
asked Dec 5 at 18:36
wondering1123
10011
1
Which part are you struggling with?
– Berci
Dec 5 at 18:44
do I ''only'' have to show, that $<f,g>$ is a scalarproduct?
– wondering1123
Dec 5 at 18:47
1
Yes. I guess the base field is $Bbb R$.
– Berci
Dec 5 at 18:49
1
@wondering1123 yes, you need to show that $langle.,.rangle_2$ is an inner product
– Masacroso
Dec 5 at 18:58
thanks! I edited some more.
– wondering1123
Dec 5 at 19:09
|
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $F$ be a surface. For all continuous functions $f,g in C(F) $ define
$$ langle f,grangle_2 := int_F f(x)g(x), dx $$
I'm struggling to show, that $ (C(F),langle.,.rangle_2) $ is a pre-hilbert space. Can you help me out? :) Any help will be very appreciated.
Showing that $<f,g>$ is a scalarproduct:
1.
$ 0 leq $ $<f,f> $
and if $ <f,f>=0 $
$ leftrightarrow int_F f(x)f(x) do(x)=0 $
then $f(x) =0 $
2.
$<f+h,g>= int_F (f+h)(x)g(x) do(x) = int_F f(x)g(x)+int_f h(x)g(x)= int_F f(x)g(x) do(x) + int h(x)g(x) do(x)= <f,g>+<h,g> $
3.
$ < lambda f,g>= lambda <f,g> $ is clear
real-analysis inner-product-space
asked Dec 5 at 18:36
wondering1123
10011
Let $F$ be a surface. For all continuous functions $f,g in C(F) $ define
$$ langle f,grangle_2 := int_F f(x)g(x), dx $$
I'm struggling to show, that $ (C(F),langle.,.rangle_2) $ is a pre-hilbert space. Can you help me out? :) Any help will be very appreciated.
Showing that $<f,g>$ is a scalarproduct:
1.
$ 0 leq $ $<f,f> $
and if $ <f,f>=0 $
$ leftrightarrow int_F f(x)f(x) do(x)=0 $
then $f(x) =0 $
2.
$<f+h,g>= int_F (f+h)(x)g(x) do(x) = int_F f(x)g(x)+int_f h(x)g(x)= int_F f(x)g(x) do(x) + int h(x)g(x) do(x)= <f,g>+<h,g> $
3.
$ < lambda f,g>= lambda <f,g> $ is clear
real-analysis inner-product-space
real-analysis inner-product-space
asked Dec 5 at 18:36
wondering1123
10011
asked Dec 5 at 18:36
wondering1123
10011
edited Dec 5 at 19:07
asked Dec 5 at 18:36
wondering1123
10011
asked Dec 5 at 18:36
wondering1123
10011
asked Dec 5 at 18:36
wondering1123
10011
10011
1
Which part are you struggling with?
– Berci
Dec 5 at 18:44
do I ''only'' have to show, that $<f,g>$ is a scalarproduct?
– wondering1123
Dec 5 at 18:47
1
Yes. I guess the base field is $Bbb R$.
– Berci
Dec 5 at 18:49
1
@wondering1123 yes, you need to show that $langle.,.rangle_2$ is an inner product
– Masacroso
Dec 5 at 18:58
thanks! I edited some more.
– wondering1123
Dec 5 at 19:09
|
show 1 more comment
1
Which part are you struggling with?
– Berci
Dec 5 at 18:44
do I ''only'' have to show, that $<f,g>$ is a scalarproduct?
– wondering1123
Dec 5 at 18:47
1
Yes. I guess the base field is $Bbb R$.
– Berci
Dec 5 at 18:49
1
@wondering1123 yes, you need to show that $langle.,.rangle_2$ is an inner product
– Masacroso
Dec 5 at 18:58
thanks! I edited some more.
– wondering1123
Dec 5 at 19:09
1
1
Which part are you struggling with?
– Berci
Dec 5 at 18:44
Which part are you struggling with?
– Berci
Dec 5 at 18:44
do I ''only'' have to show, that $<f,g>$ is a scalarproduct?
– wondering1123
Dec 5 at 18:47
do I ''only'' have to show, that $<f,g>$ is a scalarproduct?
– wondering1123
Dec 5 at 18:47
1
1
Yes. I guess the base field is $Bbb R$.
– Berci
Dec 5 at 18:49
Yes. I guess the base field is $Bbb R$.
– Berci
Dec 5 at 18:49
1
1
@wondering1123 yes, you need to show that $langle.,.rangle_2$ is an inner product
– Masacroso
Dec 5 at 18:58
@wondering1123 yes, you need to show that $langle.,.rangle_2$ is an inner product
– Masacroso
Dec 5 at 18:58
thanks! I edited some more.
– wondering1123
Dec 5 at 19:09
thanks! I edited some more.
– wondering1123
Dec 5 at 19:09
|
show 1 more comment
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1
Which part are you struggling with?
– Berci
Dec 5 at 18:44
do I ''only'' have to show, that $<f,g>$ is a scalarproduct?
– wondering1123
Dec 5 at 18:47
1
Yes. I guess the base field is $Bbb R$.
– Berci
Dec 5 at 18:49
1
@wondering1123 yes, you need to show that $langle.,.rangle_2$ is an inner product
– Masacroso
Dec 5 at 18:58
thanks! I edited some more.
– wondering1123
Dec 5 at 19:09