Approximation of a Riemann integrable function w continuous function in the L2 metric

**bubble86** · August 31st 2010, 05:44 PM

Let f be a function that is Riemann integrable. Prove there exists a piece-wise linear continuous function g such that $\ ||f-g||_2 = \sqrt (\int^b_a |f-g|^2 )$ . The given hint was construct g with a suitable partition of [b,a]. Given $\ (x_1, x_2, ...., x_n)$ define $\ g(t) = \frac{x_{i+1} - t}{\Delta x_i}\cdot f(x_i) + \frac{t-x_i}{\Delta x_i}\cdot f(x_{i+1})$ when $\ x_i\le t\le x_{i+1}$

Here is what I have so far.
So on an interval [ $\ x_i , x_{i+1}$ ] either $\ 1) f(x_i)\le g(t) \le f(x_{i+1})$ or $\ 2) f(x_{i+1})\le g(t) \le f(x_i)$
So for a given partition P $\ = (x_1, x_2, ......, x_n)$ , Let $\ M_i$ be the sup and $\ m_i$ be the inf of the interval $\ [x_i , x_{i+1}]$ .
Let h = |f-g| , for an arbitrary partition resulting from integrating f, let $\ K_i$ be the sup for the function h over the interval $\ [x_i , x_{i+1}]$ . Now $\ K_i \le \max ( {{M_i}^f} - {{m_i}^g} , {{M_i}^g} - {{m_i}^f} )$ . $\ {{M_i}^f}$ is the sup and $\ {{m_i}^f}$ is the inf of f over the interval $\ [x_i , x_{i+1}]$ . $\ {{M_i}^g}$ is the sup and $\ {{m_i}^g}$ the inf of g over the interval $\ [x_i , x_{i+1}]$ . Now $\ {{m_i}^f}\le f(x_i),f(x_{i+1}) \le {{M_i}^f}$ combined with $\ 1)$ and $\ 2)$ $\Rightarrow \ K_i \le {{M_i}^f} - {{m_i}^f}$

Because f is Riemann integrable $\exists P_\epsilon$ : $\sum_{i=1}^{n} ({{M_i}^f} - {{m_i}^f} )\cdot\Delta x_i < \epsilon$
Let $\Delta \tilde{x} = max ( \Delta x_i : 1\le i \le n)$
Once again because f is Riemann integrable $\exists P_\epsilon '$ : $\sum_{i=1}^{n'} ({{M_i}^f} - {{m_i}^f})\cdot\Delta x_i < \epsilon ' \cdot\Delta\tilde{x}$ where $\epsilon ' = \min (\epsilon , \frac{\epsilon}{\sqrt{b-a}})$
Let $\ P = P_\epsilon \bigcup P_\epsilon '$ . Construct the function g out of the partition P in piece-wise linear manner as decribed above. Let N be the number of points in P, it has at most has n + (n' -2) points and $\forall$ $\ [x_i,x_{i+1}]$ in the partition $\ P ; \Delta x_i \le\Delta \tilde{x}$ .
Let B $\ = \max ({{M_i}^f} - {{m_i}^f} : 1\le i \le N)$ , let $\ i_0$ be such a number that $\ : 1\le i_0 \le N$ and $\ B = {{M_{i_0}^f} - {{m_{i_0}^f}$
$\ {{M_{i_0}^f} - {{m_{i_0}^f}\cdot\Delta x_{i_0} \le \sum_{i=1}^{N} ({M_i}^f} - {{m_i}^f})\cdot\Delta x_i < \epsilon '\cdot\Delta\tilde{x}$
Therefore $\ B\cdot\Delta x_{i_0} < \epsilon '\cdot\Delta\tilde{x}$
$\Rightarrow B < \epsilon '$
$\sqrt (\int_b^a {|h|}^2 ) \le U({|h|}^2,P)$ $\forall$ partitions P of [a,b]. $\ U({|h|}^2,P)$ is the upper sum of the partition.
$\int_b^a {|f-g|}^2 = \int_b^a {|h|}^2 < \int_b^a {\epsilon '}^2$
$\int_b^a {\epsilon '}^2 = {\epsilon '}^2\cdot (b-a) \le (\frac{\epsilon}{\sqrt (b-a)})^2 \cdot (b-a) = \frac{{\epsilon}^2}{b-a}\cdot (b-a) = {\epsilon}^2$
$\Rightarrow ||f-g||_2 = \sqrt (\int^b_a |f-g|^2 ) < \epsilon$
Please let me know where I went wrong or if this whole approach is wrong.

can someone let me know of any good readings about Lp spaces and motivation for the Lp metric's construction.

**Jose27** · September 1st 2010, 06:08 PM

Your solution seems good (I only skimmed it though), just notice that the approach is very similar to when you try to calculate the lenght of a curve in some $\mathbb{R} ^m$ .

For the construction of the $L^p$ spaces there are several equivalent approaches (all of which require a knowledge of the Lebesgue integral): There's the measure theoretic approach, the metric space completion of certain spaces under the $L^p$ norm, etc. If you want an analytic approach Jöst's "Postmodern analysis" is a good option, if you're interested in measure theory Halmos' "Measure theory" and Bartle's "Elements of integration and Lebesgue measure" are classics.

**bubble86** · September 2nd 2010, 10:28 AM

thanks a lot for the book suggestions

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