# Limit of a complex function defined by an integral

Let $f\in L^1(\mathbb R) \cap C(\mathbb R)$, ie. $f$ is integrable and continuous. For $z \in \mathbb C$ with $Im(z) \not= 0$, define

$g(z) = \int_{-\infty}^\infty \frac{f(t)}{t-z} dt .$

I am trying to find $\lim_{Im(z) \rightarrow 0} g(z) - g(\bar z)$. I have tried writing out the expression for $g(z) - g(\bar z)$ and attempting to simplify it into a form where I can apply dominated convergence or something, but have not had much success. Any suggestions?

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There's a problem with your function : the function $g(x)$ does not depend on $x$. It is therefore constant. Do you see my point? Perhaps you meant $g(z) = \int_{-\infty}^{\infty} \frac{f(t)}{t-z} \, dt$. – Patrick Da Silva Feb 5 '12 at 3:04
This is a standard exercise on the Stieltjes transform (you are computing the Stieltjes transform of $fdm$ where $dm$ is Lebesgue). You should try contour integration. The answer is, if I recall, $\pi f(x) = \lim_{\epsilon \to 0} g(x + i\epsilon) - g(x - i \epsilon)$ There is a very nice probabilistic proof of this result as well if $f$ is the density of a probability measure. – Chris Janjigian Feb 5 '12 at 3:42
Yes, I did mean g(z). Sorry, it has been revised. – user15464 Feb 5 '12 at 3:45

Write $z = a + bi$, then $\frac{1}{(t-a)-bi}=\frac{(t-a)+bi}{(t-a)^2+b^2}$. From this we can see that $g(\bar{z}) = \overline{g(z)}$ and therefore $g(z) - g(\bar{z}) = 2\iota\Im{g(z)}$. Define $g_\epsilon(x)=\frac{\epsilon}{x^2+\epsilon^2}$. Some elementary calculus shows that $\int_\mathbb{R} g_\epsilon dm=\pi$ for all $\epsilon>0$
Now observe that $\Im g(x+i\epsilon) = \int_{-\infty}^\infty \frac{\epsilon f(t)dt}{(t-x)^2+\epsilon^2} = g_\epsilon \ast f(x)$. Since $g_\epsilon$ has a nice antiderivative when you integrate it, it is easy to show that most of its mass is concentrated around $0$ and in fact as you take $\epsilon \to 0$ all of its mass is concentrated there. For the moment,take a continuous function with compact support (you want uniform continuity) and show that for $\epsilon$ chosen sufficiently small you can force $\frac{1}{\pi}f \ast g_\epsilon (x)-f(x)$ to be arbitrarily small. I forgot a factor of $2\iota$ above. Now you have to show that you aren't too far off by taking a compact support function. To do that I think you need to split up $f$ into two parts so that you can do an infinity norm approximation near $0$ and an $L^1$ approximation with Young's lemma near infinity.