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.