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

Here is what I have so far.

So on an interval [$\displaystyle \ x_i , x_{i+1}$] either $\displaystyle \ 1) f(x_i)\le g(t) \le f(x_{i+1})$ or $\displaystyle \ 2) f(x_{i+1})\le g(t) \le f(x_i)$

So for a given partition P $\displaystyle \ = (x_1, x_2, ......, x_n)$, Let $\displaystyle \ M_i$ be the sup and $\displaystyle \ m_i$ be the inf of the interval $\displaystyle \ [x_i , x_{i+1}]$.

Let h = |f-g| , for an arbitrary partition resulting from integrating f, let $\displaystyle \ K_i$ be the sup for the function h over the interval $\displaystyle \ [x_i , x_{i+1}]$ . Now $\displaystyle \ K_i \le \max ( {{M_i}^f} - {{m_i}^g} , {{M_i}^g} - {{m_i}^f} )$. $\displaystyle \ {{M_i}^f}$ is the sup and $\displaystyle \ {{m_i}^f}$ is the inf of f over the interval $\displaystyle \ [x_i , x_{i+1}]$. $\displaystyle \ {{M_i}^g}$ is the sup and $\displaystyle \ {{m_i}^g}$ the inf of g over the interval $\displaystyle \ [x_i , x_{i+1}]$. Now $\displaystyle \ {{m_i}^f}\le f(x_i),f(x_{i+1}) \le {{M_i}^f}$ combined with $\displaystyle \ 1)$ and $\displaystyle \ 2)$ $\displaystyle \Rightarrow \ K_i \le {{M_i}^f} - {{m_i}^f}$

Because f is Riemann integrable $\displaystyle \exists P_\epsilon$ : $\displaystyle \sum_{i=1}^{n} ({{M_i}^f} - {{m_i}^f} )\cdot\Delta x_i < \epsilon$

Let $\displaystyle \Delta \tilde{x} = max ( \Delta x_i : 1\le i \le n)$

Once again because f is Riemann integrable $\displaystyle \exists P_\epsilon '$ : $\displaystyle \sum_{i=1}^{n'} ({{M_i}^f} - {{m_i}^f})\cdot\Delta x_i < \epsilon ' \cdot\Delta\tilde{x}$ where $\displaystyle \epsilon ' = \min (\epsilon , \frac{\epsilon}{\sqrt{b-a}})$

Let $\displaystyle \ P = P_\epsilon \bigcup P_\epsilon '$. 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 $\displaystyle \forall $ $\displaystyle \ [x_i,x_{i+1}] $ in the partition $\displaystyle \ P ; \Delta x_i \le\Delta \tilde{x}$.

Let B $\displaystyle \ = \max ({{M_i}^f} - {{m_i}^f} : 1\le i \le N)$, let $\displaystyle \ i_0 $ be such a number that $\displaystyle \ : 1\le i_0 \le N $ and $\displaystyle \ B = {{M_{i_0}^f} - {{m_{i_0}^f}$

$\displaystyle \ {{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 $\displaystyle \ B\cdot\Delta x_{i_0} < \epsilon '\cdot\Delta\tilde{x} $

$\displaystyle \Rightarrow B < \epsilon '$

$\displaystyle \sqrt (\int_b^a {|h|}^2 ) \le U({|h|}^2,P) $ $\displaystyle \forall$partitions P of [a,b]. $\displaystyle \ U({|h|}^2,P)$ is the upper sum of the partition.

$\displaystyle \int_b^a {|f-g|}^2 = \int_b^a {|h|}^2 < \int_b^a {\epsilon '}^2$

$\displaystyle \int_b^a {\epsilon '}^2 = {\epsilon '}^2\cdot (b-a) \le (\frac{\epsilon}{\sqrt (b-a)})^2 \cdot (b-a) = \frac{{\epsilon}^2}{b-a}\cdot (b-a) = {\epsilon}^2$

$\displaystyle \Rightarrow ||f-g||_2 = \sqrt (\int^b_a |f-g|^2 ) < \epsilon$

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.