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Math Help - Principal value

  1. #1
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    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
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  2. #2
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    Re: Principal value

    I would assume this is true, from my very loose argument that $\displaystyle \begin{align*} \int_a^a{f(x)\,\mathrm{d}x} = 0 \end{align*}$.
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    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<K, 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.
    Last edited by SlipEternal; June 16th 2014 at 07:12 PM.
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    Re: Principal value

    Quote Originally Posted by SlipEternal View Post
    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<K, 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
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    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<t<\delta, 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.
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    Re: Principal value

    Quote Originally Posted by SlipEternal View Post
    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<t<\delta, 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
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