Let $\alpha$ be a fixed increasing function on $[a,b]$. For $u\in \mathcal{R}(\alpha)$, define $$\lVert u \rVert_2=\left(\int \limits_{a}^{b}|u|^2d\alpha\right)^{1/2}.$$

Suppose $f\in \mathcal{R}(\alpha)$ and $\epsilon>0$. Prove that there exists a continuous function $L$ on $[a,b]$ such that $\lVert f-L \rVert_2<\epsilon.$

Proof: Let $\epsilon>0$ be given. Since $f\in \mathcal{R}(\alpha)$ then exists partition $P=\{x_0,x_1,\cdots,x_n\}$ of $[a,b]$ such that $U(P,f,\alpha)-L(P,f,\alpha)<\dfrac{\epsilon^2}{2M}$ where $M=\sup\limits_{[a,b]}|f|$. Let $L(t)$ be polygonal chain connecting $\{f(x_0), f(x_1),\cdots, f(x_n)\}$. Hence $$L(t)=\dfrac{x_i-t}{\Delta x_i}f(x_{i-1})+\dfrac{t-x_{i-1}}{\Delta x_i}f(x_{i})$$ for $t\in [x_{i-1},x_i]$. Then $$\lVert f-L \rVert_2^2=\int \limits_{a}^{b}|f-L|^2d\alpha=\sum \limits_{i=1}^{n}\int \limits_{x_{i-1}}^{x_{i}}|f-L|^2d\alpha\leqslant\sum \limits_{i=1}^{n}\sup \limits_{[x_{i-1},x_i]}|f-L|^2\Delta\alpha_i.$$ It's easy to check that $|f-L|\leqslant M_i-m_i$ since $L(x_{i-1})=f(x_{i-1})$ and $L(x_{i})=f(x_{i})$. Hence $$\sup \limits_{[x_{i-1},x_i]}|f-L|^2\leqslant (M_i-m_i)^2=(M_i-m_i)(M_i-m_i)=$$$$=\left(\sup \limits_{[x_{i-1},x_i]}f-\inf \limits_{[x_{i-1},x_i]}f\right)(M_i-m_i)=\sup\limits_{[x_{i-1},x_i]}|f(x)-f(x')|(M_i-m_i)\leqslant 2M_i(M_i-m_i)\leqslant 2M(M_i-m_i).$$ Thus $$\lVert f-L \rVert_2^2\leqslant 2M\sum \limits_{i=1}^{n}(M_i-m_i)\Delta\alpha_i=2M(U(P,f,\alpha)-L(P,f,\alpha))<\epsilon^2.$$ Hence we get that $$\lVert f-L \rVert_2<\epsilon$$

What do you think about my proof? Is it correct?

  • $\begingroup$ It's been quite a while since I studied Stieljes integration, but I do not believe that $f$ being $\alpha$-itegrable is enough to show it bounded, even on a compact set. If so, then your line $M = \sup\limits_{[a,b]} |f|$ is not justified. $\endgroup$ – Paul Sinclair Nov 13 '15 at 1:15
  • $\begingroup$ @PaulSinclair, You think that my proof is not correct? Which moment you think is incorrect? P.S. $f$ being $\alpha$-integrable we suppose that $f$ is bounded (it's definition by W.Rudin). $\endgroup$ – ZFR Nov 13 '15 at 5:46
  • $\begingroup$ I just told you exactly what I noticed wrong with it. Why are you asking me to repeat it? I was unaware that Rudin defines Stieljes integrability only for bounded functions. That is a rather limiting definition, even if it is useful for you here. It is certainly not universal. Are you sure he defines it that way? $\endgroup$ – Paul Sinclair Nov 13 '15 at 14:42
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    $\begingroup$ Next questions: what are $M_i$ and $m_i$, that appear in your proof without previous definition? (Max and Min of $f$ on the i-th interval, based on how you relate them to the upper and lower sums later, but you should state that instead of leaving people to deduce it). But if $m_i < 0$, then $(M_i - m_i)^2 \le M_i^2 - m_i^2$ does not hold. $\endgroup$ – Paul Sinclair Nov 13 '15 at 17:16
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    $\begingroup$ Obviously so. Your proof works as is on non-negative $f$, because then $m_i \ge 0$ everywhere. In the general case, since you know $f$ is bounded, there is some $K$ (I'm changing from $M$, since you already use $M$) such that $f + K$ is non-negative. If $L$ approximates $f + K$, then $L - K$ approximates $f$. $\endgroup$ – Paul Sinclair Nov 13 '15 at 19:45

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