# Math Help - Principal value

1. ## Principal value

I couldn't find anything on the net that dealt with this, which probably means I have the wrong term, but just a quick check.

A function $\psi (x)$ is finite and continuous at the origin. Can we then say that
$\lim_{\epsilon \to 0} \int_{- \epsilon}^{\epsilon} \psi ~ dx = 0$

Thanks!
-Dan

2. ## Re: Principal value

I would assume this is true, from my very loose argument that \displaystyle \begin{align*} \int_a^a{f(x)\,\mathrm{d}x} = 0 \end{align*}.

3. ## Re: Principal value

That depends... there are functions that are continuous at a point, but not Riemann integrable on any interval containing that point. Are you saying $\psi$ is Riemann integrable on some interval containing zero? If so, then yes, the limit is zero. Here is a quick sketch of a proof:

Let $m_\epsilon = \inf\{\psi(x)\mid -\epsilon \le x \le \epsilon\}, M_\epsilon = \sup\{\psi(x)\mid -\epsilon \le x \le \epsilon\}$.

Then $2\epsilon m_\epsilon\le \int_{-\epsilon}^\epsilon \psi(x)dx \le 2\epsilon M_\epsilon$.

You can easily show that given $0 < \epsilon_1 < \epsilon_2$, $m_{\epsilon_2} \le m_{\epsilon_1} \le M_{\epsilon_1} \le M_{\epsilon_2}$. Hence, there exists $K>0$ and $m,M\in \Bbb{R}$ such that for all $0<\epsilon, you have $m \le m_\epsilon$ and $M_\epsilon \le M$. Thus, $2\epsilon m\le\int_{-\epsilon}^\epsilon \psi(x)dx \le 2\epsilon M$ for all $0< \epsilon < K$. Then, by the Squeeze Theorem, you have your result.

4. ## Re: Principal value

Originally Posted by SlipEternal
That depends... there are functions that are continuous at a point, but not Riemann integrable on any interval containing that point. Are you saying $\psi$ is Riemann integrable on some interval containing zero? If so, then yes, the limit is zero. Here is a quick sketch of a proof:

Let $m_\epsilon = \inf\{\psi(x)\mid -\epsilon \le x \le \epsilon\}, M_\epsilon = \sup\{\psi(x)\mid -\epsilon \le x \le \epsilon\}$.

Then $2\epsilon m_\epsilon\le \int_{-\epsilon}^\epsilon \psi(x)dx \le 2\epsilon M_\epsilon$.

You can easily show that given $0 < \epsilon_1 < \epsilon_2$, $m_{\epsilon_2} \le m_{\epsilon_1} \le M_{\epsilon_1} \le M_{\epsilon_2}$. Hence, there exists $K>0$ and $m,M\in \Bbb{R}$ such that for all $0<\epsilon, you have $m \le m_\epsilon$ and $M_\epsilon \le M$. Thus, $2\epsilon m\le\int_{-\epsilon}^\epsilon \psi(x)dx \le 2\epsilon M$ for all $0< \epsilon < K$. Then, by the Squeeze Theorem, you have your result.
A "quick" sketch? I admit that when I read this the first thing out of my head was "WTF??"

The proof is beyond me. I'll settle for the results for now!

-Dan

5. ## Re: Principal value

Have you ever seen the construction for Volterra's function? Is is a function that is everywhere differentiable, its derivative is bounded everywhere, but this derivative is not Riemann-integrable. You might be able to use a similar construction to find a function that is continuous at zero, but not Riemann-integrable on any interval containing zero. You just need the function to be discontinuous on a set of positive measure in every open interval containing zero, yet still continuous at zero.

Anyway, you said that $\psi(x)$ was finite, but you did not say it was bounded. So, I actually need to revise the "sketch" for the proof. I was using the assumption that it was bounded. Since it is continuous at zero, given any $\epsilon>0$, there exists $\delta>0$ such that for all $|x| < \delta$, $|\psi(x)-\psi(0)|<\epsilon$. Hence, $\psi(0)-\epsilon < \psi(x) < \psi(0)+\epsilon$ for any $|x|<\delta$. This follows from the definition of continuity.

So, let's choose $\delta>0$ such that for all $-\delta < x < \delta$, we have $\psi(0)-1 < \psi(x) < \psi(0)+1$. The Riemann integral is supposed to give the area under a curve. That area is bounded by the minimum and maximum height of the function. Since we know that $\psi(x)$ will never go below $\psi(0)-1$, that is the lower bound for the value of the function. We know it will never go above $\psi(0)+1$. This is true on any subinterval of $(-\delta,\delta)$. I will replace the $\epsilon$ that you are using with $t$ so as not to confuse it with the $\epsilon=1$ we used for continuity to find $\delta>0$. So, for any $0, the rectangle with base from $-t$ to $t$ has a length of $2t$. It has a width (or height) of at minimum $\psi(0)-1$ and at maximum $\psi(0)+1$. So, $2t(\psi(0)-1) \le \int_{-t}^t \psi(x)dx \le 2t(\psi(0)+1)$ (since the integral in the middle is a sum of areas of rectangles, and we know none of those rectangles can have heights outside of that range). Since $\psi(0)-1$ is finite and $\psi(0)+1$ is finite, we have:

$\lim_{t \to 0} 2t(\psi(0)-1) = 0 = \lim_{t\to 0} 2t(\psi(0)+1)$

Then, by the Squeeze Theorem, we have $\lim_{t\to 0} \int_{-t}^t \psi(x)dx = 0$.

Again, all of this depends on $\psi(x)$ being Riemann-integrable on some open interval containing zero.

6. ## Re: Principal value

Originally Posted by SlipEternal
Have you ever seen the construction for Volterra's function? Is is a function that is everywhere differentiable, its derivative is bounded everywhere, but this derivative is not Riemann-integrable. You might be able to use a similar construction to find a function that is continuous at zero, but not Riemann-integrable on any interval containing zero. You just need the function to be discontinuous on a set of positive measure in every open interval containing zero, yet still continuous at zero.

Anyway, you said that $\psi(x)$ was finite, but you did not say it was bounded. So, I actually need to revise the "sketch" for the proof. I was using the assumption that it was bounded. Since it is continuous at zero, given any $\epsilon>0$, there exists $\delta>0$ such that for all $|x| < \delta$, $|\psi(x)-\psi(0)|<\epsilon$. Hence, $\psi(0)-\epsilon < \psi(x) < \psi(0)+\epsilon$ for any $|x|<\delta$. This follows from the definition of continuity.

So, let's choose $\delta>0$ such that for all $-\delta < x < \delta$, we have $\psi(0)-1 < \psi(x) < \psi(0)+1$. The Riemann integral is supposed to give the area under a curve. That area is bounded by the minimum and maximum height of the function. Since we know that $\psi(x)$ will never go below $\psi(0)-1$, that is the lower bound for the value of the function. We know it will never go above $\psi(0)+1$. This is true on any subinterval of $(-\delta,\delta)$. I will replace the $\epsilon$ that you are using with $t$ so as not to confuse it with the $\epsilon=1$ we used for continuity to find $\delta>0$. So, for any $0, the rectangle with base from $-t$ to $t$ has a length of $2t$. It has a width (or height) of at minimum $\psi(0)-1$ and at maximum $\psi(0)+1$. So, $2t(\psi(0)-1) \le \int_{-t}^t \psi(x)dx \le 2t(\psi(0)+1)$ (since the integral in the middle is a sum of areas of rectangles, and we know none of those rectangles can have heights outside of that range). Since $\psi(0)-1$ is finite and $\psi(0)+1$ is finite, we have:

$\lim_{t \to 0} 2t(\psi(0)-1) = 0 = \lim_{t\to 0} 2t(\psi(0)+1)$

Then, by the Squeeze Theorem, we have $\lim_{t\to 0} \int_{-t}^t \psi(x)dx = 0$.

Again, all of this depends on $\psi(x)$ being Riemann-integrable on some open interval containing zero.
This time I understood it. Yay! Yes, by finite I meant bounded (above and below.) Thanks for the efforts, I appreciate them.

-Dan