Exercise 6.12

Question

\def\eps{\varepsilon}
Exercise 12: With the notations of Exercise 11, suppose $f\in\mathscr R(\alpha)$ and $\varepsilon>0$. Prove that there exists a continuous function $g$ on $[a,b]$ such that
    $||f-g||_2<\varepsilon$.

analambanomenos · Accepted Answer

\def\eps{\varepsilon}

Since $f\in\mathscr R(\alpha)$, There are bounds $m\le f(x)\le M$ for the values of $f$ in $[a,b]$. Let $\varepsilon>0$ and let $P=\{x_0,\ldots,x_n\}$ be a partition of
    $[a,b]$ such that $U(f,P,\alpha)-L(f,P,\alpha)\le\varepsilon^2/(M-m)$. Following the hint, for $t\in[x_{i-1},x_i]$ define $$g(t)=\frac{x_i-t}{\Delta x_i}f(x_{i-1})+
    \frac{t-x_{i-1}}{\Delta x_i}f(x_i).$$ Then $g$ is linear on $[x_{i-1},x_i]$ and the definitions of $g$ on $[x_{i-1},x_i]$ and $[x_i,x_{i+1}]$ both define $g(x_i)=f(x_i)$,
    and so $g$ is a continuous function on $[a,b]$. Also, $m_i\le g(t)\le M_i$ on each $[x_{i-1},x_i]$. Hence 
    \begin{align*}
        ||f-g||_2^2 &= \int_a^b|f-g|^2\,d\alpha \
        &= \sum_{i=0}^n\int_{x_{i-1}}^{x_i}|f-g|^2\,d\alpha \
        &\le \sum_{i=0}^n(M_i-m_i)^2\Delta\alpha_i \
        &\le (M-m)\sum_{i=0}^n(M_i-m_i)\Delta\alpha_i \
        &= (M-m)\bigl(U(f,P,\alpha)-L(f,P,\alpha)\bigr) \
        &\le \varepsilon^2
    \end{align*}

Exercise 6.12

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Exercise 6.12

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