Regulated Function

**Showcase_22** · April 3rd 2010, 09:40 AM

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

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

Here's my answer.

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

Claim: $\phi_n$ is odd.

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

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

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

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

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

We can now equate the two to get:

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

So we know that $\phi_n$ is odd.

My next claim is:

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

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

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

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

We can then rewrite this as:

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

Is this the right?

**Defunkt** · April 3rd 2010, 01:52 PM

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

**Showcase_22** · April 3rd 2010, 11:04 PM

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.

**Drexel28** · April 4th 2010, 08:11 PM

Originally Posted by Showcase_22

H

Claim: $\phi_n$ is odd.

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

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

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

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

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

We can now equate the two to get:

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

So we know that $\phi_n$ 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 $0=f(x)+f(-x)=\lim_{n\to\infty}\left(\phi_n(x)+\phi_n(-x)\right)$ and thus eventually $\phi_n(x)=-\phi_n(-x)$ . I don't see why that implies that $\phi_n(x)$ is odd for all $n$ .

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

**Showcase_22** · April 5th 2010, 03:12 AM

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

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

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

Therefore:

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

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

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

But since f is odd:

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

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

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

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

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