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I'm currently trying to prove the Riemann-Lebesgue lemma using lower Darboux-sums and an approximation of any integrable function $f: [0,1] \to \mathbb{R}$ defined as $$t(x) := \begin{cases} m_i & \text{for } x_i \leq x < x_{i+1},\; i < n\\ m_n & \text{for } x= 1 \end{cases},$$ where $m_i := \inf\nolimits_{\xi \in [x_i, x_{i+1}]} f(\xi)$ and $P = \{x_i\}_{1 \leq i \leq n}$ is a partition of the interval $[0,1]$. Now I want to show that $$\int_0^1 t(x) \cos(\lambda x) \; \mathrm dx \leq \varepsilon$$ for some $\varepsilon > 0$.

Our assistant professor told us that it was possible to actually integrate $t(x) \cos(\lambda x)$ but I just don't see how to. I guess I would have to make use of the product rule twice (as $\int t(x) \; \mathrm dx = s(f,P)$), but the problem is that I don't know the integral of the lower Darboux-sum (or whether it even exists) and also I don't know the derivative of $t(x)$ (or whether it exists).

How am I supposed to proceed? Is there a different trick to integrate $t(x) \cos(\lambda x)$?

Thanks for any answers in advance.

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    $\begingroup$ Try to do the integral by splitting $[0,1]$ into $[x_0,x_1], [x_1,x_2],\cdots$. This way, on each interval $t(x)$ is a constant, and $\cos$ is something you know how to integrate. $\endgroup$
    – user325
    Commented Apr 14, 2011 at 20:55
  • $\begingroup$ $t(x)$ is piecewise constant $\endgroup$
    – yoyo
    Commented Apr 14, 2011 at 20:56
  • $\begingroup$ Integrating piecewise, I arrive at $\frac{1}{\lambda} \sum_{l=1}^{n-1} m_l \cdot (\sin(\lambda x_{l+1}) - \sin(\lambda x_l))$. Can I now simply estimate $\sum_l m_l \leq n \sup f(x)$ and then for all $\lambda \geq \lambda_0 = 2 n \sup f(x)/\varepsilon$ the integral is less or equal to $\varepsilon$? I ask this, because in the solution the integral is basically the same as mine but contains $(x_{i+1} - x_i)$ instead of $m_l$, where of course $\sum_i x_{i+1} - x_i = 1$ which would be lots nicer. $\endgroup$
    – Huy
    Commented Apr 14, 2011 at 21:17
  • $\begingroup$ let $\lambda\to\infty$ in $(1/\lambda)\sum_l(\sin(\lambda x_{l+1})-\sin(\lambda x_l))$. dont get hung up on details; the point is that this gets small as $\lambda$ gets big $\endgroup$
    – yoyo
    Commented Apr 15, 2011 at 1:22

1 Answer 1

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We can integrate piecewise:

$\int_0^1 t(x) \cos(\lambda x) \; \mathrm dx = \sum_{i=1}^{n-1} \int_{x_i}^{x_{i+1}} m_i \cos(\lambda x) \; \mathrm dx = \sum_{i=1}^{n-1} m_i \cdot \int_{x_i}^{x_{i+1}} \cos(\lambda x) \; \mathrm dx$

$ = \sum_{i=1}^{n-1} m_i \cdot \left. \left( \frac{1}{\lambda}\cdot \sin(\lambda x) \right) \right|_{x_i}^{x_{i+1}}$

$ = \frac{1}{\lambda} \cdot \sum_{i=1}^{n-1} m_i \cdot (\sin(\lambda x_{i+1}) - \sin(\lambda x_i)).$

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