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I'm trying to do the following exercise from Rudin's Real and Complex Analysis:

Suppose $f\in C(T)$ and $f\in \text{Lip }\alpha$ for some $\alpha>0$. Prove that the Fourier series of $f$ converges to $f(x)$, by completing the following outline: It is enough to consider the case $x=0$, $f(0)=0$. The difference between the partial sums $s_n(f;0)$ and the integrals $$\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\frac{\sin nt}{t}\:dt$$ tends to $0$ as $n\to\infty$. The function $f(t)/t$ is in $L^1(T)$. Apply the Riemann-Lebesgue lemma. More careful reasoning shows that the convergence is actually uniform on $T$.

I already know how to prove the result as in Lipschitz Continuity and Hölder Continuity helps Fourier series to converge.

However, I don't understand Rudin's outline.

  1. When he says that it is enough to consider the case $x=0$, $f(0)=0$, is it simply because it changes almost nothing to the proof?
  2. Since we can prove that $s_n(f;0)$ tends to $0$ by proving that $f(t)\cot(t/2)$ is in $L^1$, it seems to me that Rudin's outline is only better if it is easier to prove that $f(t)/t\in L^1$ and that the difference between $s_n(f;0)$ and the integrals tends to $0$. But I don't see any easy way to do the latter.
  3. Why is the convergence uniform and why does it matter?

I would appreciate if anyone could answer those 3 questions.

Gabriel
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  • The first point seems nothing more than a "without loss of generality" argument. You can rotate in $\mathbb{T}$ just by adding a phase and add a constant without changing the convergence of the Fourier sequence. – Adrián González Pérez Jul 30 '19 at 10:12
  • @AdriánGonzález-Pérez Hm, ok. If $x=a$, we can just study $g(t-a)$ instead. Also we can add constants without changing the convergence. Is this that you mean? – Gabriel Jul 30 '19 at 12:42

2 Answers2

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  1. The aim is to show that $s_n(f,a) \rightarrow f(a) $ for all $a$, however it is easy to show that if you let $g(x)=g(x+a)-g(a)$, $$s_n(f,a)-s_n(g,0) = \sum_{-n}^n \frac1{2\pi}\int_{-\pi}^\pi (f(t)e^{ik(a-t) } - g(t)e^{-ikt})dt = f(a)$$ so if you showed that $s_n(g,0)\rightarrow 0$ you may deduce that $s_n(f,a)\rightarrow f(a)$.

  2. I think I agree with what you say. As $$s_n(f,0)=\frac1{2\pi}\int_{-\pi}^\pi f(t)\left(\sum_{-n}^n e^{-ikt}\right)dt = \frac1{2\pi}\int_{-\pi}^\pi f(t)\left(\sin(nt)\cot(t/2)+\cos(nt)\right)dt.$$ There is no additional complexity in showing that $t\mapsto f(t)\cot(t/2)$ is $L^1$ compared to $t\mapsto f(t)/t$, and then the result comes directly from Riemann-Lebesgue lemma. My only guess is that Rudin might be implicitly referring to the approximation of the Dirichlet kernel by the sinus cardinal function: $$D_n(x)=\frac{2\sin (nx)}x+\left(\sin (nx) \left(\operatorname{cotan} \frac x 2-\frac2x\right)+\cos(nx)\right)$$ where $D_n(t)=\sum_{-n}^n e^{-ikt}$.

  3. Uniform convergence is much stronger than pointwise convergence, it definitely makes a difference!

FXV
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  • Do you have an idea about how could we prove that the difference between $s_n(f,0)$ and the integrals from the statement tend to $0$? – Gabriel Aug 08 '19 at 09:44
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    You could want to use an approximation of the Dirichlet kernel by the sinus cardinal function: $D_n(x)=\frac{2\sin (nx)}x+\left(\sin (nx) \left(\operatorname{cotan} \frac x 2-\frac2x\right)+\cos(nx)\right)$. But that would still mean applying Riemann-Lebesgue lemma twice! – FXV Aug 08 '19 at 19:44
  • Yeah, this works nicely! I just don't yet understand why the convergence is uniform. I tried to use Arzèla-Ascoli to prove it but couldn't prove that $(s_n(f;x))$ is equicontinuous. – Gabriel Aug 09 '19 at 18:08
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    I think you can show uniform equicontinuity by splitting carefully the integral formulation of $|s_n(f,x)-f(x)-(s_n(f,y)-f(y))|$. Indeed you get something like $\int_{-\pi}^\pi D_n(t)[f(x-t)-f(x)-(f(y-t)-f(y))]dt$. When $|t|<|x-y|$ you can bound $|f(x-t)-f(x)|$ by $C|t|^\alpha$ and $|f(y-t)-f(y)|$ likewise. When $|t|>|x-y|$ you can pair the terms differently and bound $|f(x-t)-f(y-t)|$ by $C|x-y|^\alpha$ and $|f(x)-f(y)|$ likewise. Conjugated to the fact that $|D_n(t)|<K|t|^{-1}$, you get the uniform equicontinuity! – FXV Aug 10 '19 at 19:21
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Since it may be helpful to someone, here is my complete answer to this question (done, of course, with a lot of help from @FXV).

Firstly, if $g(t)=f(t+x)-f(x)$, then by linearity we have that $$s_n(f;x)-s_n(g;0)=f(x).$$ In other words, if $s_n(g;0)\to 0$, then $s_n(f;x)\to f(x)$ so it suffices to consider the case $x=0$, $f(0)=0$.

Now we use the usual formula for $\sin(x+y)$ to write $$D_n(t)=\frac{\sin (n+\frac{1}{2})t}{\sin(t/2)}=2\cdot\frac{\sin nt}{t}+\left[\sin nt\left(\cot (t/2)-\frac{2}{t}\right)+\cos nt\right].$$ Since $D_n$ is even, this implies that the difference in the exercise's statement is equal to $$\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)\left(\cot (t/2)-\frac{2}{t}\right)\sin nt\:dt+\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)\cos nt\:dt.$$ But $\cot(t/2)-2/t\to 0$ as $t\to 0$, which implies that this function has only a removable singularity. Thus, by defining it to be $0$ for $t=0$ we see that both $$f(t)\left(\cot (t/2)-\frac{2}{t}\right) \quad\text{and}\quad f(t)$$ are continuous functions and belong to $L^1(T)$. The Riemann-Lebesgue lemma then implies that both integrals tend to $0$ as $n\to\infty$.

Since $f\in \text{Lip }\alpha$, $$\left|\frac{f(t)}{t}\right|=\frac{|f(t)-f(0)|}{|t-0|}\leq M_f |t-0|^{\alpha-1}=M_f|t|^{\alpha-1}$$ for all $t\neq 0$. This implies that $|f(t)/t|$ is integrable and so $f(t)/t\in L^1(T)$. Finally, as we saw, $$s_n(f;0)=\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\frac{\sin nt}{t}\:dt+x_n,$$ where $x_n\to 0$ as $n\to \infty$. A third application of the Riemann-Lebesgue lemma implies that $s_n(f;0)\to 0$.

In view of Arzelà-Ascoli's theorem (theorem 7.25 in Principles of Mathematical Analysis), it suffices to show that $\{s_n(f;x)-f(x)\}$ is equicontinuous to have uniform convergence. For that, we have to bound $$|[s_n(f;x)-f(x)]-[s_n(f;y)-f(y)]|.$$ Since $\int_{-\pi}^\pi D_n(t)\:dt=2\pi$, this is equal to $$\left|\int_{-\pi}^\pi \left\{[f(x-t)-f(x)]-[f(y-t)-f(y)]\right\} D_n(t)\:dt\right|.$$ Using the triangular inequality twice we bound this integral by \begin{multline*} \int_A \left\{ |f(x-t)-f(x)|+|f(y-t)-f(y)| \right\} |D_n(t)|\:dt+\\ \int_B \left\{ |f(x-t)-f(y-t)|+|f(x)-f(y)| \right\} |D_n(t)|\:dt, \end{multline*} where $A=\{t\in [-\pi,\pi]\::\: |t|<|x-y|\}$ and $B=\{t\in [-\pi,\pi]\::\: |t|>|x-y|\}$. As $f\in \text{Lip }\alpha$, it follows that this is bounded by $$\int_A \left\{2M_f |t|^\alpha\right\} |D_n(t)|\:dt+\int_B \left\{2M_f |x-y|^\alpha\right\}|D_n(t)|\:dt.$$ Now, since $(t/2)/\sin(t/2)\to 1$ as $t\to 0$, there exists $\delta>0$ such that $$|D_n(t)|<4|t|^{-1}$$ for all $0<|t|<\delta$. We conclude that, if $|x-y|<\min(\delta,\pi)$, the integral over $A$ is bounded by $$8M_f\int_A|t|^{\alpha-1}\:dt$$ and that the integral over $B$ is bounded by $$2M_f|x-y|^{\alpha} \underbrace{\int_B |D_n(t)|\:dt}_{\text{bounded}}.$$ Finally, since $-1<\alpha-1\leq 0$, we can estimate $\int_A |t|^{\alpha-1}\:dt$ in the following way: let $|x-y|=a$ and observe that \begin{align*} \int_A |t|^{\alpha-1}\:dt &= 2 \int_0^{a} t^{\alpha-1}\:dt\\ &= 2\left(\int_{a/2}^{a} t^{\alpha-1}\:dt+\int_{a/4}^{a/2} t^{\alpha-1}\:dt+\dotsc \right)\\ &\leq 2\left(\frac{a}{2}\left(\frac{a}{2}\right)^{\alpha-1}+\frac{a}{4}\left(\frac{a}{4}\right)^{\alpha-1}+\dotsc\right)\\ &=K|x-y|^{\alpha}, \end{align*} for some constant $K>0$. Putting it all together, we get that if $|x-y|<\min(\delta,\pi)$, then $$|[s_n(f;x)-f(x)]-[s_n(f;y)-f(y)]|<K'|x-y|^\alpha,$$ for another constant $K'>0$. This implies that $\{s_n(f;x)-f(x)\}$ is equicontinuous and the result follows.

Gabriel
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  • Thanks for the detailed proof, just be careful with the integral of $D_n$ on $B$: It is not bounded by a constant but it is $O(\ln(|x-y|))$ so it can be managed. – FXV Aug 11 '19 at 16:38
  • @FXV Since $|D_n(t)|\to 0$ for all $t\in B$, so does the integral (by DTC). In other words, there exists $N$ such that $\int_B |D_n(t)|<1$ for all $n>N$. But $D_n(t)$ is in $L^1(B)$ for all $n$, so we can bound uniformly $\int_B |D_n(t)|$ by the maximum of this (finite) value for $n\leq N$ and $1$. Isn't this correct? – Gabriel Aug 12 '19 at 11:41
  • The problem is that $B$ depends on $|x-y|$, which you want to take as small as required to show the equicontinuity. – FXV Aug 12 '19 at 11:52
  • @FXV Makes sense. How could we fix it? My method doesn't seem to allow an explicit bound. – Gabriel Aug 12 '19 at 11:58
  • Yes it does because you still have $|D_n(t)|<4|t|^{-1}$. So $8M_f\int_A|t|^{\alpha-1}:dt$ can be bounded by $16M_f\alpha^{-1}|x-y|^\alpha$ and $2M_f|x-y|^{\alpha} \int_B |D_n(t)|:dt$ by $16M_f|x-y|^{\alpha} (\ln(\pi)-\ln(|x-y|)):dt$. Both bounds are $o(|x-y|)$. – FXV Aug 12 '19 at 19:09
  • But nothing guarantees that $\delta\geq \pi$ and so $|D_n(t)|<4|t|^{-1}$. – Gabriel Aug 12 '19 at 19:18
  • You can use the concavity of $\sin$ over $[0,\pi/2]$ to show that $\sin(t)\geq 2t/\pi$ for all $t\in[0,\pi/2]$. Thus $|D_n(t)|\leq \pi |t|^{-1}$. This makes your constant even smaller (not that it matters much). – FXV Aug 12 '19 at 19:53
  • Sorry @Gabriel, could you explain me in detail why $s_n(f; x) - s_n(g;0) = f(x)$? It is not so clear for me right now. – C. Bishop Apr 09 '20 at 16:43
  • @C.Bishop Did you try writing out the integrals? It is just computation, nothing else. – stoic-santiago Jul 20 '21 at 07:14