Verify property of inner product

[psie]

Homework Statement

Let ##w(t)## be a nonnegative continuous function on ##I=[a,b]## such that if ##f\in C(I)## and ##\int_I w(t)f(t) dt=0##, then ##f=0##. Prove that $$\langle f,g\rangle=\int_a^b w(t)f(t) \overline{g(t)} dt,$$ defines an inner product on ##L^2(I)##.

Relevant Equations

The properties of an inner product; conjugate symmetry, linearity in the first argument and positive-definiteness.

I struggle with verifying positive-definiteness, in particular $$\langle f,f\rangle =0\implies f=0.$$ I know that for continuous non-negative functions, if the integral vanishes, then the function is identically ##0##. Here, however, ##f## being in ##L^2## does not make it continuous, right? This is from Bridge's Foundations of Real and Abstract Analysis.

 

[fresh_42]

That definition is confusing. Why should we care about continuity if all that counts is Lebesgue integrability?

I have a version in my book where everything is about continuity (not ##L_2##), and a version about Lebesgue integrability (##L_2##) without weight function ##w.## Your version looks like comparing apples and oranges to me.

E.g., we have by the mean value theorem that
$$
0=\bigl\langle f\, , \,f \bigr\rangle =\int_a^b w(t) \|f\|^2 \,dt = w(\xi)\int_a^b \|f\|^2 \,dt
$$
And here is the problem: if ##w(\xi)\neq 0## then we are done. But we cannot know that this is the case. If we remain in the box of oranges, and demand
$$
\int_a^b w(t)h(t)\,dt =0 \Longrightarrow h\equiv 0 \text{ for all } h\in L_2(I)
$$
instead of restricting us to continuous functions only, then it makes sense and we can conclude by setting ##h=\|f\|^2.## The other possibility is to require ##w## to be strictly positive. That would fill the gaps of potential Lebesgue Null sets, too.

 

[andrewkirk]

Homework Statement: Let ##w(t)## be a nonnegative continuous function on ##I=[a,b]## such that if ##f\in C(I)## and ##\int_I w(t)f(t) dt=0##, then ##f=0##. Prove that $$\langle f,g\rangle=\int_a^b w(t)f(t) \overline{g(t)} dt,$$ defines an inner product on ##L^2(I)##.
Relevant Equations: The properties of an inner product; conjugate symmetry, linearity in the first argument and positive-definiteness.

I struggle with verifying positive-definiteness, in particular $$\langle f,f\rangle =0\implies f=0.$$ I know that for continuous non-negative functions, if the integral vanishes, then the function is identically ##0##. Here, however, ##f## being in ##L^2## does not make it continuous, right? This is from Bridge's Foundations of Real and Abstract Analysis.

According to Wolfram, ##L^2## is actually not a set of functions, but of equivalence classes of functions, where functions that are identical except on sets of measure zero are equivalent.
Under that interpretation, we can conclude from ##\langle f, f\rangle = 0## that ##f## is zero except on a set of measure zero. That means that ##f## is in the same equivalence class as the zero function and hence is zero in the space ##L^2##.

 

[fresh_42]

According to Wolfram, ##L^2## is actually not a set of functions, but of equivalence classes of functions, where functions that are identical except on sets of measure zero are equivalent.
Under that interpretation, we can conclude from ##\langle f, f\rangle = 0## that ##f## is zero except on a set of measure zero. That means that ##f## is in the same equivalence class as the zero function and hence is zero in the space ##L^2##.

That's the case if ##w\equiv 1.## But how do we exclude that ##L_2\ni w\equiv 0.##?

 

[andrewkirk]

That's the case if ##w\equiv 1.## But how do we exclude that ##L_2\ni w\equiv 0.##?

It is not ##w## but ##f## and ##g## that are indicated to be elements of ##L^2## in the OP.
##w## is the measure function used to define the inner product on the space. In the wolfram article I linked they call it ##\mu##.
If ##w\equiv 0## then all sets have ##w##-measure zero, so there is only one equivalence class of functions, and the inner product space becomes trivial.

 

[fresh_42]

It is not ##w## but ##f## and ##g## that are indicated to be elements of ##L^2## in the OP.
##w## is the measure function used to define the inner product on the space. In the wolfram article I linked they call it ##\mu##.
If ##w\equiv 0## then all sets have ##w##-measure zero, so there is only one equivalence class of functions, and the inner product space becomes trivial.

It remains to show that ##\bigl\langle f\, , \,f \bigr\rangle = 0 ## implies ##f=0.## How is it done?

We must somehow rule out that ##w(t)=0## whenever ##f(t)\neq 0.## I see this if we require
a) ##w(t)>0## for all ##t\in I##
or
b) ##\int_I w(t)f(t)dt=0 \Rightarrow f=0## for all ##f\in L_2##

I do not see it if ##f\in C(I)## in condition b), a significantly smaller set of functions.

 

[fresh_42]

Just a remark.

The continuity requirement for the weight function ##w## makes sense because it defines a property for all ##f\in L_2## and not those depending on ##w.## The weight shouldn't disturb, and continuity guarantees that.

 

[PeroK]

Homework Statement: Let ##w(t)## be a nonnegative continuous function on ##I=[a,b]## such that if ##f\in C(I)## and ##\int_I w(t)f(t) dt=0##, then ##f=0##.

Isn't that impossible? Whatever ##w## you choose, can't you find some non-zero ##f## that makes the integral vanish?

 

[PeroK]

Homework Statement: Let ##w(t)## be a nonnegative continuous function on ##I=[a,b]## such that if ##f\in C(I)## and ##\int_I w(t)f(t) dt=0##, then ##f=0##.

Here's an outline proof that any such ##w \equiv 0## on ##[a, b]##

As ##w## is continuous and nonnegative on ##[a, b]##, then either ##w \equiv 0## or ##\int_a^b w(t) dt > 0##. Assume the latter.

We can find ##a < c < b##, such that:
$$\int_a^c w(t)dt = \int_c^b w(t) dt > 0$$Now, choose any continuous, non-negative function ##f_1## defined on ##[a, c]## such that ##f_1(c) = 0##. And any continuous non-positive function ##f_2## defined on ##[c, b]## such that ##f_2(c) = 0##. Also, neither ##f_1## nor ##f_2## is identically zero.

Now, we have:
$$\int_a^c w(t)f_1(t)dt > 0, \ \text{and} \ \int_c^b w(t)f_2(t) dt < 0$$Note that we can multiply ##f_2## by some constant so that:
$$\int_a^c w(t)f_1(t)dt = -\int_c^b w(t)f_2(t) dt$$Finally, the function ##f## defined as ##f_1## on ##[a, c]## and ##f_2## on ##[c, b]## is continuous, not identially zero and satisfies:
$$\int_a^b w(t)f(t)dt = 0$$Hence, there is no ##w## with the originally stated property.

 

[psie]

The other possibility is to require ##w## to be strictly positive. That would fill the gaps of potential Lebesgue Null sets, too.

I know the assumption are quite confusing, probably not right either, but why would a positive weight function make the statement ##\langle f,f\rangle\implies f=0## a.e. true?

 

[martinbn]

I know the assumption are quite confusing, probably not right either, but why would a positive weight function make the statement ##\langle f,f\rangle\implies f=0## a.e. true?

Do you mean ##\langle f, f\rangle = 0##? That implies that ##w(t)|f(t)|^2=0##, which means ##f=0## a.e.

 

[martinbn]

Is this from a textbook?

 

[fresh_42]

Is this from a textbook?

This is from Bridge's Foundations of Real and Abstract Analysis.

I assume that it is a cut-and-paste error or a simple little mistake.

 

[martinbn]

Isn't that impossible? Whatever ##w## you choose, can't you find some non-zero ##f## that makes the integral vanish?

This is strange. I looked in the book and it is written exactly like that.

 

[PeroK]

This is strange. I looked in the book and it is written exactly like that.

Although I laboured through a proof, it's obvious there can be no such ##w##.

 

[fresh_42]

Although I laboured through a proof, it's obvious there can be no such ##w##.

Matches with my textbook: ##w\ge 0## in the case of ##C(I)## and ##w\equiv 1## for ##L_2(I)##.

 

[PeroK]

Matches with my textbook: ##w\ge 0## in the case of ##C(I)## and ##w\equiv 1## for ##L_2(I)##.

The initial property is close to saying that the only continuous function whose integral is zero on a closed interval is the zero function. That's what it amounts to.

 

[fresh_42]

The initial property is close to saying that the only continuous function whose integral is zero on a closed interval is the zero function. That's what it amounts to.

Well, I haven't found that strange condition, only the definitions of inner products on the two spaces.

 

[martinbn]

Although I laboured through a proof, it's obvious there can be no such ##w##.

No, i agree with you. It is strange that it is in the book.

 

[martinbn]

May be ##C(I)## means continuous and positive. And positive means non-negative. This way ##w## is continuous non-negative and is zero on set of measure zero.

 

[Bosko]

Homework Statement: Let ##w(t)## be a nonnegative continuous function on ##I=[a,b]## such that if ##f\in C(I)## and ##\int_I w(t)f(t) dt=0##, then ##f=0##.

Consider functions in the form ##f_c(t)=c-t##

For ##c=a##, function ##f_{a}(t)=a-t \leq 0## , for all ##t \in [a,b]##.
For ##c=b+1##, function ##f_{b+1}(t)=b+1-t \geq 1## , for all ##t \in [a,b]##.

Function
$$F(c)=\int_a^b w(t) (c-t) dt$$
is continuous. For ##F(a) \leq 0## and ##F(b+1) \geq 0##.

There exists ##c_0 \in [a, b+1]## such that
$$F(c_0)=\int_a^b w(t) (c_0-t) dt=0$$
but ##f(t)=c_0-t## is not 0 for all ##t \in [a,b]##

I assume that ##f(t)## should also be a nonnegative continuous function.

Here's an outline proof that any such ##w \equiv 0## on ##[a, b]##
...
$$\int_a^b w(t)f(t)dt = 0$$Hence, there is no ##w## with the originally stated property.

Now I see that I did the same thing in a similar way.

 

[mathwonk]

hmmm,... it seems as if there is no such function w, even if a=b.