1. ## Regulated Function

Let $\displaystyle f:[-1,1] \rightarrow \mathbb{R}$ be an odd regulated function. Prove that:

$\displaystyle \int_{-1}^1f=0$.

By definition of a regulated function $\displaystyle \int_{-1}^1f= \lim_{n \rightarrow \infty} \phi_n$ where $\displaystyle \phi_n$ is a sequence of step functions converging uniformly to f.

Claim: $\displaystyle \phi_n$ is odd.

We already know that $\displaystyle \lim_{n \rightarrow \infty} \phi_n(x)=f(x)$. (*)

We can multiply both sides by -1 to get $\displaystyle - \lim_{n \rightarrow \infty} \phi_n(x)=-f(x)$ for some $\displaystyle x \in [-1,1]$.

Since (*) is valid $\displaystyle \forall \ x \in [-1,1]$, we also know that:

$\displaystyle \lim_{n \rightarrow \infty}\phi_n(-x)=f(-x)$.

But we can rewrite the first one as $\displaystyle - \lim_{n \rightarrow \infty} \phi_n(x)=-f(x)=f(-x)$ since we know that f is odd.

We can now equate the two to get:

$\displaystyle - \lim_{n \rightarrow \infty} \phi_n(x)=\lim_{n \rightarrow \infty} \phi_n(-x)$

So we know that $\displaystyle \phi_n$ is odd.

My next claim is:

$\displaystyle \phi_k:[-1,1] \rightarrow \mathbb{R}$ is an odd step function so $\displaystyle \int_{-1}^1 \phi_k=0$ where $\displaystyle k=0,1, \ldots$.

$\displaystyle \int_{-1}^1 \phi_k==\int_{-1}^0 \phi_k+\int_0^1 \phi_k=\int_{-1}^0 \phi_k(x)~dx+\int_0^1 \phi_k(x)~dx$.

We can use the fact $\displaystyle \phi_k$ is odd:

$\displaystyle =\int_{-1}^0 \phi_k(x)~dx-\int_0^1 - \phi_k(x)~dx=\int_{-1}^0 \phi_k(x)~dx-\int_0^1 -\phi_k(-x)~dx$

We can then rewrite this as:

$\displaystyle =\int_{-1}^0 \phi_k(x)~dx-\int_{-1}^0 \phi_k(x)~dx=0$

Is this the right?

2. Isn't $\displaystyle \int_{-a}^a f(x)dx = 0$ for all odd functions? Why do you need it to be regulated?

3. I don't really think it's that necessary for it to be regulated, it's just what the question (and my course) is about.

I imagine it's a case of proving existing theories from a regulated/step function perspective.

4. Originally Posted by Showcase_22
H

Claim: $\displaystyle \phi_n$ is odd.

We already know that $\displaystyle \lim_{n \rightarrow \infty} \phi_n(x)=f(x)$. (*)

We can multiply both sides by -1 to get $\displaystyle - \lim_{n \rightarrow \infty} \phi_n(x)=-f(x)$ for some $\displaystyle x \in [-1,1]$.

Since (*) is valid $\displaystyle \forall \ x \in [-1,1]$, we also know that:

$\displaystyle \lim_{n \rightarrow \infty}\phi_n(-x)=f(-x)$.

But we can rewrite the first one as $\displaystyle - \lim_{n \rightarrow \infty} \phi_n(x)=-f(x)=f(-x)$ since we know that f is odd.

We can now equate the two to get:

$\displaystyle - \lim_{n \rightarrow \infty} \phi_n(x)=\lim_{n \rightarrow \infty} \phi_n(-x)$

So we know that $\displaystyle \phi_n$ is odd.
I agree that the regulated part (I haven't even heard this term used except rarely in books on Banach spaces!) is overkill. But, for the above, why can't you say that $\displaystyle 0=f(x)+f(-x)=\lim_{n\to\infty}\left(\phi_n(x)+\phi_n(-x)\right)$ and thus eventually $\displaystyle \phi_n(x)=-\phi_n(-x)$. I don't see why that implies that $\displaystyle \phi_n(x)$ is odd for all $\displaystyle n$.

For example $\displaystyle \phi_n(x)=\begin{cases} \frac{x}{n} & \mbox{if} n\geqslant 10^{10^{10}} \\ \pi & \mbox{if} \quad n<10^{10^{10}}\end{cases}$

5. That's actually very true! I've come up with something else:

Claim: $\displaystyle \phi_p, \ 1 \leq p \leq n$ is odd.

By definition of a regulated function, we can choose $\displaystyle \phi_p$, as defined above, s.t $\displaystyle |\phi_p(x)-f(x)| < \frac{\epsilon}{2}$ where $\displaystyle x \in [-1,1]$.

Therefore:

$\displaystyle -\frac{\epsilon}{2}<\phi_p(x)-f(x)<\frac{\epsilon}{2}$ (*)

Since $\displaystyle x \in [-1,1]$, and $\displaystyle \phi_p(x)$ and $\displaystyle f(x)$ are defined at all values of $\displaystyle x$:

$\displaystyle -\frac{\epsilon}{2}<\phi_p(-x)-f(-x)<\frac{\epsilon}{2}$

But since f is odd:

$\displaystyle -\frac{\epsilon}{2}<\phi_p(-x)+f(x)<\frac{\epsilon}{2}$ (**)

So we can add (*) and (**) together to get:

$\displaystyle -\epsilon<\phi_p(x)+\phi_p(-x)<\epsilon$

and this implies that $\displaystyle \phi_p(x)=-\phi_p(-x)$ or, in a more conventional way, $\displaystyle -\phi_p(x)=\phi_p(-x)$