MHB Does This Lemma Prove All Continuous Functions Orthogonal to $C^2$ Are Zero?

[evinda]

Hello! (Wave)

Lemma:

If $f \in C([a,b])$ and $\int_a^b f(x) h(x) dx=0 \ \forall h \in C^2([a,b])$ with $h(a)=h(b)=0$ then $f(x)=0 \ \forall x \in [a,b]$.

Proof of lemma:

Suppose that there is a $x_0 \in (a,b)$ such that $f(x_0) \neq 0$, for example without loss of generality we suppose that $f(x_0)>0$.
Because of continuity there is an interval $[x_1, x_2] \subset (a,b)$ such that $x_0 \in (x_1, x_2)$ and $f(x)>0 \ \forall x \in (x_1, x_2)$.

We define the function $g(x)=\left\{\begin{matrix}
(x_2-x)^3 (x-x_1)^3 & , x \in (x_1, x_2)\\ \\
0 & , x \in [a,b] \setminus{(x_1,x_2)}
\end{matrix}\right.$.

Then $g \in C^2([a,b])$ and $g(a)=g(b)=0$. From the hypothesis we have:

$$\int_a^b f(x)g(x) dx=0$$

But $\int_a^b f(x)g(x) dx= \int_{x_1}^{x_2} f(x)g(x) dx>0$, contradiction.First of all, why do we say that there is an interval $[x_1, x_2] \subset (a,b)$ such that $x_0 \in (x_1, x_2)$ and $f(x)>0 \forall x \in (x_1, x_2)$? Why don't we pick the closed interval $[x_1, x_2]$ ? (Thinking)

Also why does it hold that $\int_a^b f(x) g(x) dx= \int_{x_1}^{x_2} f(x) g(x) dx$?

Furthermore, the prof told us that we couldn't take the function

$g(x)=\left\{\begin{matrix}
(x_2-x)^2 (x-x_1)^2 & , x \in (x_1, x_2)\\ \\
0 & , x \in [a,b] \setminus{(x_1,x_2)}
\end{matrix}\right.$

but the powers both of $(x_2-x), (x-x_1)$ have to be greater or equal to $3$. Why is it like that? (Thinking)

 

[GJA]

Hi evinda,

I will try to answer each of the questions you've asked in order:

Hello! (Wave)

Lemma:

If $f \in C([a,b])$ and $\int_a^b f(x) h(x) dx=0 \ \forall h \in C^2([a,b])$ with $h(a)=h(b)=0$ then $f(x)=0 \ \forall x \in [a,b]$.

Proof of lemma:

Suppose that there is a $x_0 \in (a,b)$ such that $f(x_0) \neq 0$, for example without loss of generality we suppose that $f(x_0)>0$.
Because of continuity there is an interval $[x_1, x_2] \subset (a,b)$ such that $x_0 \in (x_1, x_2)$ and $f(x)>0 \ \forall x \in (x_1, x_2)$.

We define the function $g(x)=\left\{\begin{matrix}
(x_2-x)^3 (x-x_1)^3 & , x \in (x_1, x_2)\\ \\
0 & , x \in [a,b] \setminus{(x_1,x_2)}
\end{matrix}\right.$.

Then $g \in C^2([a,b])$ and $g(a)=g(b)=0$. From the hypothesis we have:

$$\int_a^b f(x)g(x) dx=0$$

But $\int_a^b f(x)g(x) dx= \int_{x_1}^{x_2} f(x)g(x) dx>0$, contradiction.First of all, why do we say that there is an interval $[x_1, x_2] \subset (a,b)$ such that $x_0 \in (x_1, x_2)$ and $f(x)>0 \forall x \in (x_1, x_2)$? Why don't we pick the closed interval $[x_1, x_2]$ ? (Thinking)

This question really gets at the nitty gritty, but I will do my best to answer it.

KEY IDEA: It is vital that $f>0$ on a nonempty open interval for the part of the proof where the integral
$$\int_{x_{1}}^{x_{2}}f(x)g(x)~dx$$
comes in, because nonempty open intervals are "intervals" in the intuitive sense; i.e. they have a positive length/are not single points.

Suppose that your professor's proof did not mention at all an open interval; i.e. suppose it read exactly the same way but with the open interval replaced with a closed interval. Then it would say:

Because of continuity there is an interval $[x_1, x_2] \subset (a,b)$ such that $x_0 \in [x_1, x_2]$ and $f(x)>0 \ \forall x \in [x_1, x_2]$.

There are two different reasons this is bad:

1) The statement could be flat out false. It is possible that one of the endpoints, say $x_{1},$ is a zero of $f.$ Then we couldn't say $f>0$ for all $x\in [x_{1},x_{2}].$ Ultimately, though, this would not kill the proof so long as $x_{1}<x_{2}$ and $f>0$ on $(x_{1},x_{2}).$ I only mention all of this for the sake of being complete in my response.

Note: 1) above has nothing to do with 2) below.

2) Read the "new" statement with the closed interval once again. A litigious reader would say that you technically have not mentioned anything to preclude the case where $x_{1}=x_{0}=x_{2}.$ This would completely destroy the "KEY IDEA" mentioned above, because you are now integrating over a single point. The result of this integration would be zero, and now you can't derive the contradiction that you once were able to with your professor's argument.

Note: I say "litigious reader" above because a more moderate reader would most likely understand the introductory clause "Because of continuity" to mean $x_{1}<x_{2}.$ At any rate, the best way to write the proof is how your instructor has, because it avoids all this tedium. Ultimately this comes down communicating mathematics in the most efficient, correct way.

Also why does it hold that $\int_a^b f(x) g(x) dx= \int_{x_1}^{x_2} f(x) g(x) dx$?

Because $g(x)=0$ outside of the interval $[x_{1},x_{2}].$

Furthermore, the prof told us that we couldn't take the function

$g(x)=\left\{\begin{matrix}
(x_2-x)^2 (x-x_1)^2 & , x \in (x_1, x_2)\\ \\
0 & , x \in [a,b] \setminus{(x_1,x_2)}
\end{matrix}\right.$

but the powers both of $(x_2-x), (x-x_1)$ have to be greater or equal to $3$. Why is it like that? (Thinking)

Recall that $g$ needs to be $C^{2},$ which means that $g''$ must be continuous. Take two derivatives of the polynomial part of $g$ and note that if you plug in $x_{1}$ that the result is not $0.$ Thus $g''$ is not continuous. Having a third power avoids this problem.

Let me know if anything is unclear/not quite right.