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This is based on an approach to a homework problem from last year that I discarded but would like to fill in the details of now. Unfortunately I can't see the necessary inequality.

Suppose we are given a function $g$ of bounded variation on $[a,b]$. I would like to construct the Lebesgue-Stieltjes integral with respect to $g$ by defining a bounded linear functional on $C^1([a,b])$ by $\varphi \mapsto -\int_{\mathbb{R}} g(x) \varphi^{'}(x) dx$ then extending it to $C([a,b])$ by density and then applying the Riesz Representation Theorem. To do this, however, I need to show that $|\int g(x) \varphi^{'}(x) dx|\leq C||\varphi||_\infty$ (notice that there is no derivative here--this is the condition to make it a bounded functional on $C^0([a,b])$) which is where the trouble comes in since I am not sure how to get such a bound without integrating by parts (which we can't do since $g$ need not be absolutely continuous).

I'm sure there is a nice way to handle this but I just don't see it.

Thanks for the help.

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Once you know that $g\in L^1$, you can employ Holder's inequality:'s_inequality – William Feb 12 '12 at 20:55
Can you elaborate a bit? I am not sure how to get a bound using the infinity norm of $\varphi$ using Holder's inequality on $g$ and $\varphi^{'}$. – Chris Janjigian Feb 12 '12 at 20:57
Holder's inequality holds for the conjugate pair $p = 1$ and $q = \infty$ (see the link I provided). Now, if $g\in L^1$, for any $\psi\in L^\infty$, you have: $|\int g(x)\psi(x)dx|\leq \int|g(x)\psi(x)|dx\leq \|\psi\|_\infty\int|g(x)|dx$. Now set $C = \int|g(x)|dx = \|g\|_{1}$. – William Feb 12 '12 at 21:00
Right but I need the bound in terms of the infinity norm of $\varphi$ and we have the derivative of $\varphi$ in the integral here. What you are saying gives me a bound of $||g||_1||\varphi^{'}||_\infty$, not $C||\varphi||_\infty$ which is what is necessary for this to be a bounded linear functional on $C^0$. – Chris Janjigian Feb 12 '12 at 21:01
@Chris: Very good. I missed that, thanks for point that out. – William Feb 12 '12 at 21:05
up vote 1 down vote accepted

Since you are working on $C^1([a,b])$, you can make use of the usual Riemann-Stieltjes integral to see that

$$ \begin{align*} \int_a^b \phi(x) \, dg(x) &= \left[\phi(b)g(b) - \phi(a)g(a)\right] - \int_a^b g(x) \, d\phi(x) \\ &= \left[\phi(b)g(b) - \phi(a)g(a)\right] - \int_a^b g(x)\phi'(x) \, dx \end{align*} $$

And using $\left|\int_a^b \phi(x) \, dg(x) \right|\le \|\phi\|_{\infty} \; V_a^b(g)$ (where $V_a^b(g)$ denotes the total variation of $g$), we obtain

$$\left|\int_a^b g(x)\phi'(x) \, dx\right| \le \left[|g(a)|+|g(b)|+V_a^b(g)\right] \, \| \phi\|_\infty = C\cdot \|\phi\|_\infty$$

So your functional is indeed continuous (and the Lebesgue-Stieltjes integral is an extension of the Riemann-Stieltjes integral).

In case you are not familiar with the Riemann-Stieltjes integral. The first equality can be seen as follows: Given a partition $P = \{x_0,x_1, \dots, x_N\}$ of $[a,b]$; $a= x_0 < x_1 <\dots < x_N=b$ and $\xi_i \in [x_{i-1}, x_{i}]$, we can write

$$ \begin{align} \sum_{n=1}^N f(\xi_i) [g(x_i) - g(x_{i-1})] &= f(\xi_{N+1})g(b) - f(\xi_1)g(a) - \sum_{n=1}^N \left[f(\xi_{i+1}) - f(\xi_i)\right] g(x_{i}) \\ &= f(\xi_{N+1})g(b) - f(\xi_1)g(a) - \sum_{n=1}^N \frac{f(\xi_{i+1}) - f(\xi_i)}{\xi_{i+1} - \xi_i} g(x_{i}) \left[\xi_{i+1} - \xi_i\right] \end{align} $$

setting $\xi_{N+1} = b$. Note that $$\left|\sum_{n=1}^N f(\xi_i) \left[g(x_i) - g(x_{i-1})\right]\right| \le \sum_{n=1}^N \left|f(\xi_i)\right| \, \left|g(x_i) - g(x_{i-1})\right|\le \|f\|_\infty \, V_a^b(g)$$ for any partition $P$.

So if $f \in C^1([a,b])$ and $g$ is of bounded variation, then both are Riemann-integrable and by taking the limit $|P|\to 0$ (finesse of the partition) we will obtain the estimate

$$\left| f(b)g(b) - f(a)g(a) - \int_a^b f'(x) g(x) \, dx \right| \le \|f\|_\infty \, V_a^b(g)$$

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I don't think $f$ as defined here has to be absolutely continuous. Can you explain why the Cantor function is not a counterexample here? – Chris Janjigian Feb 13 '12 at 23:19
@Chris: You are absolutely right. I was under the impression that "bounded variation + continuity" implied "absolute continuity" (I haven't thought about Stieltjes integrals in a while, sorry). But this is not the case, as the Cantor function shows. – Sam Feb 14 '12 at 2:32
@Chris: I have corrected it now. – Sam Feb 14 '12 at 2:59
That is a very nice solution indeed, thank you! I should have added though that I was hoping to find a way to do this without going through the Riemann-Stieltjes integral. This originated as part of a question from Rudin Real and Complex which, as an intermediate step, includes the construction of the Lebesgue-Stieltjes integral. The way I solved it was to just define the Riemann-Stieltjes integral as a functional. The problem is that Rudin does not do this in R&C and so I am trying to figure out how to define a functional equivalent to Riemann-Stieltjes without using its existence. – Chris Janjigian Feb 14 '12 at 3:28
@Chris: Hmm, this seems not so easy. Have you already thought about first defining a map $g\mapsto \Lambda_g$ from some suitable dense subspace of BV (like $\{g = f+h\}$, where $f$ is absolutely continuous and $h$ is constant up to jumps, maybe) into the space of linear functionals on $C^1([a,b])$, e.g. by $\Lambda_g(\phi) = \int_a^b \phi(x) f'(x) \, dx + \sum_{x} \phi(x)[h(x+) - h(x-)]$ and proving that this (linear) mapping is continuous; so you could first extend $g\mapsto \Lambda_g$ to all of BV and then - for fixed $g$ of bounded variation - extend $\Lambda_g$ to all of $C^1([a,b])$. – Sam Feb 14 '12 at 4:51

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