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Show $\int_{0}^{1}{g(t)\over t-z}dt$, $g(t):[0,1]\to \Bbb{R}$ a continuous function, is holomorphic in $\Bbb{C}\setminus[0,1]$. Trying to simplify ${F(z+h)-F(z)\over h}$ I arrived at $\int_{0}^{1}{g(t)\over (t-z)(t-(z+h))}dt$, and I don't know how I better compute its limit as $h\to 0$. Can I put the limit inside? How? Following Calculus, I only know I can do this for $n$ functions (countable ones) uniformly converging to $F$ but this is not the case. In addition, I don't know how to relate to $[0,1]$ points in this process so at to truly be sure how they can or can't impact the convergence of the limit above. I would appreciate your help.

Edit: I added more tags mentioning Real Analysis because I don't want to ignore the probability that one whose expertise is Real Analysis and not Complex Analysis will be able to give a Real-Analysis-based explanation

Meitar
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  • What is $g$ ..........? – Empty Nov 04 '15 at 15:50
  • You have miscopied or miscalculated. $$\frac{1}{t - (z+h)} - \frac{1}{t-z} = \frac{h}{(t - (z+h))(t-z)},$$ so your denominator ought to be $(t-z)(t-(z+h))$, not $z(z+h)$. – Daniel Fischer Nov 04 '15 at 15:52
  • I am so sorry I forgot to mention it. I edited my question. – Meitar Nov 04 '15 at 15:52
  • Daniel, I think you are correct. Let me check it again. If what I wrote really were the case, I wouldn't be having this problem. – Meitar Nov 04 '15 at 15:53
  • That doesn't affect the argument, however. In any case, we have uniform convergence of the integrand as $h \to 0$, so you can bring the limit inside the integral. – Daniel Fischer Nov 04 '15 at 15:55

2 Answers2

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HINT:

$$\begin{align} \left|\dfrac{F(z+h)-F(z)}{h}-\int_0^1 \frac{g(t)}{(t-z)^2}\,dt\right|&=\left|\frac1h\int_0^1 g(t)\left(\dfrac{1}{t-(z+h)}-\dfrac{1}{t-z}-\frac{h}{(t-z)^2}\right)\,dt\right|\\\\ &=\left|\int_0^1 g(t)\left(\dfrac{1}{(t-z)(t-z-h)}-\frac{1}{(t-z)^2}\right)\,dt\right|\\\\ &=\left|\int_0^1 \dfrac{hg(t)}{(t-z)^2(t-z-h)}\,dt\right|\\\\ &\le\int_0^1 \dfrac{|h|\,|g(t)|}{|t-z|^2|t-z-h|}\,dt\\\\ \end{align}$$

SPOILER ALERT Scroll over the highlighted area to reveal the solution

We note that $|t-z-h|=\sqrt{(t-h)^2-2\text{Re}(z)(t-h)+|z|^2}\ge \left|\text{Im}(z)\right|$. Therefore, we have $$\int_0^1\left|\dfrac{hg(t)}{(t-z)^2(t-z-h)}\right|\,dt\le \dfrac{h}{\left|\text{Im}(z)\right|}\int_0^1\dfrac{|g(t)|}{(t-z)^2}\,dt \tag 1$$Since the integrand in $(1)$ is bounded and independent of $h$, then it is easy to see that $$\lim_{h\to 0}\int_0^1\left|\dfrac{hg(t)}{(t-z)^2(t-z-h)}\right|\,dt=0$$And this completes the proof that $F(z)$ is differentiable.

Mark Viola
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  • Thank you for simplifying. What I find harder is the actual computing of the limit. – Meitar Nov 04 '15 at 16:05
  • I added the solution in a "Spoiler Alert" section. – Mark Viola Nov 04 '15 at 16:21
  • Thank you for elaborating. Can I ask why absolute value is needed? Does it prove differentiability without the absolute value? – Meitar Nov 04 '15 at 17:29
  • You're welcome. Here, the "absolute value" is the complex magnitude. The development establishes that the upper bound for the magnitude (which is non-negative) can be made arbitrarily small (i.e., less than any pre-assigned $\epsilon>0$) by choosing $h$ sufficiently small. That means that the limit of the difference quotient approaches the integral of $\int_0^1 g(t)\dfrac{d}{dz}\left(\dfrac{1}{t-z}\right),dt$. So, yes. This proves that $F$ is differentiable for all $z$ in the stated domain. – Mark Viola Nov 04 '15 at 17:56
  • I could really use elaboration on the note hidden above. – Meitar Nov 05 '15 at 15:32
  • I'm happy to help. So, please help me understand where you would like clarification/elucidation. – Mark Viola Nov 05 '15 at 17:44
  • I am still learning complex analysis so I may be incorrect, but I believe the first inequality in the hint is incorrect. For instance, fixing $t\in[0,1]$ we can just set $z=t+ri$ for some real $r>0$, so that $z$ is positioned directly above $t$ in the plane. Then if we add $h=si$ for some small negative real number $s$, we see that $z+h$ is slightly below $z$ in the plane, and the vertical distance $|t-(z+h)|$ will be less than the vertical distance $|t-z|=|\Im(z)|$. I think changing $\Im(z)$ to $\Im(z+h)$ should work and the rest of the argument can be easily adjusted . – modz Feb 04 '25 at 21:49
  • @modz Note that $$\begin{align} |z-(t-h)|^2&=|z|^2+(t-h)^2-2(t-h)\text{Re}(z)\\ &=\left(\text{Im}(z)\right)^2 +\left(\text{Re}(z)-(t-h)\right)^2\\ &\ge\left(\text{Im}(z)\right)^2 \end{align}$$So, the development is sound – Mark Viola Feb 04 '25 at 23:16
  • (+1) Nice and concise. Years later after your posting I answered this elated question about the Stieltjes transform here. I'd appreciate your commenting on it. – Mittens Feb 05 '25 at 01:12
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HINT: Please double check your result of ${F(z+h)-F(z)\over h}$ $${F(z+h)-F(z)\over h}=\frac{1}{h}(\int_{0}^{1}{g(t)\over t-(z+h)}dt-\int_{0}^{1}{g(t)\over t-z}dt)$$

$${F(z+h)-F(z)\over h}=\frac{1}{h}(\int_{0}^{1}{[(t-z)-(t-z-h)] g(t)\over (t-(z+h))(t-z)}dt)$$ $${F(z+h)-F(z)\over h}=\frac{1}{h}(\int_{0}^{1}{h g(t)\over (t-(z+h))(t-z)}dt)$$

$$\lim\limits_{ h\to 0 }{ {F(z+h)-F(z)\over h} }= \lim\limits_{ h\to 0 }{ \frac{1}{h}(\int_{0}^{1}{h g(t)\over (t-(z+h))(t-z)}dt)}$$

$$\lim\limits_{ h\to 0 }{ {F(z+h)-F(z)\over h} }= \lim\limits_{ h\to 0 }{ \frac{h}{h}(\int_{0}^{1}{ g(t)\over (t-(z+h))(t-z)}dt)}$$

$$\lim\limits_{ h\to 0 }{ {F(z+h)-F(z)\over h} }= \lim\limits_{ h\to 0 }{ (\int_{0}^{1}{ g(t)\over (t-(z+h))(t-z)}dt)}=\int_{0}^{1}{ g(t)\over (t-z)^2}dt$$

Mathlover
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  • Yes I just arrived at it. My main problem is the handling of this limit. – Meitar Nov 04 '15 at 15:56
  • What justifies moving the limit inside the integral? This is what I couldn't explain. – Meitar Nov 04 '15 at 16:01
  • The integral depends on $t$ but we take limit on $h$. – Mathlover Nov 04 '15 at 16:07
  • Yes that's what I thought, but still, there's got to be a formal way to justify it. Regardless of this specific case, because after all, it's not the same function as $h$ changes, so what keeps it legitimate? Continuity? I have been looking for formal theorem and restrictions and haven't come by something to unequivocally support this procedure. – Meitar Nov 04 '15 at 16:13
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    Please check Leibniz integral rule. The proof has similiar idea. https://en.wikipedia.org/wiki/Leibniz_integral_rule#Proofs Then have a look the title 'General form with variable limits' in the page – Mathlover Nov 04 '15 at 17:21