Example 5
Given the series \[ f(\theta)=\sum_{n=1}^{\infty}\frac{1}{2^n}\sin(n\theta). \] Is this a Fourier series of a continuous function?
If it's continuous, is it differentiable?

First note that here we are given the function \(f\) defined above. In other words,

$$ \begin{align*} f(\theta) := \lim_{N\to\infty}S_N(\theta), \end{align*} $$

where

$$ \begin{align*} S_N(\theta) = \sum_{n=1}^{N}\frac{1}{2^n}\sin(n\theta). \end{align*} $$
So the question is: is the function obtained by summing this infinite series continuous?
$$ \begin{align*} \sum_{n=1}^{\infty}\frac{1}{2^n}\sin(n\theta) \end{align*} $$

This is different from the previous examples where we’re given a “proposed” Fourier series for \(f\) and we’re trying to determine whether it is actually \(f\). In that case, we would define partial sum \(S_N = \sum_{n=1}^{N}\widehat f(n) e^{inx}\) and then

  1. Assuming \(f\) is Reimann integrable, we would compute the Fourier coefficient of \(f\)
  2. Use the \(M\)-test to show that the sequence of partial sums \(S_N\) converges uniformly to some function \(g\). Since each \(S_N\) is continuous, their uniform limit \(g\) is also continuous.
  3. By uniform convergence, we get to interchange the sum and integral when we expand the definition of \(\widehat g\). Hence we get to integrate term by term to show that the Fourier coefficients of \(g\) are exactly the coefficients of \(f\) that we computed
  4. By the Uniqueness Theorem of Fourier series, we then can show that \(f = g\) almost everywhere.
  5. Since \(g\) is continuous, if \(f\) is also continuous, then equality almost everywhere improves to equality everywhere: that is, \(f=g\).

Is \(f\) continuous?

In this particular example, \(f\) is defined to be

$$ \begin{align*} f(\theta)=\sum_{n=1}^{\infty}\frac{1}{2^n}\sin(n\theta) \end{align*} $$

Hence this isn’t a proposal. We know it’s \(f\). Now, let

$$ \begin{align*} S_N(\theta)=\sum_{n=1}^{N}\frac{1}{2^n}\sin(n\theta). \end{align*} $$

Each \(S_N\) is continuous. Also observe that

$$ \begin{align*} \left|\frac{1}{2^n}\sin(n\theta)\right| \leq \frac{1}{2^n} \end{align*} $$

for every \(\theta\), hence in the Weierstrass M-test, we choose \(M_n = \frac{1}{2^n}\). Now note that

$$ \begin{align*} \sum_{n=1}^{\infty}M_n = \sum_{n=1}^{\infty}\frac{1}{2^n} =\frac{\frac{1}{2}}{1-\frac{1}{2}} = 1<\infty. \end{align*} $$

Each \(S_N\) is continuous because it is a finite sum of continuous functions. The Weierstrass \(M\)-test shows that the series converges absolutely for every \(\theta\), and that its sequence of partial sums \(S_N\) converges uniformly to \(f\).”

$$ \begin{align*} f(\theta) = \sum_{n=1}^{\infty}\frac{1}{2^n}\sin(n\theta). \end{align*} $$

Finally, by the uniform limit theorem, since each \(S_N\) is continuous and the sum converges uniformly, then \(f\) must be continuous.

Is \(f\) differentiable?

At this point, we know that the sequence of partial sums

$$ \begin{align*} S_N(\theta) = \sum_{n=1}^{N}\frac{1}{2^n}\sin(n\theta), \end{align*} $$

converges uniformly to \(f\). Moreover, we know that each partial sum is differentiable since it’s a finite sum of differentiable functions. Then if we can prove that the derivative series

$$ \begin{align*} \sum_{n=1}^{\infty} \left(\frac{1}{2^n}\sin(n\theta) \right)' \end{align*} $$

converges uniformly, then the term-by-term differentiation theorem tells us that \(f\) is differentiable and

$$ \begin{align*} f'(\theta) &= \sum_{n=1}^{\infty} \left(\frac{1}{2^n}\sin(n\theta)\right)' = \sum_{n=1}^{\infty}\frac{n}{2^n}\cos(n\theta). \end{align*} $$

So let’s define the partial sum

$$ \begin{align*} \boxed{ S_N'(\theta) = \left(\frac{1}{2^n}\sin(n\theta)\right)' = \sum_{n=1}^{N}\frac{n}{2^n}\cos(n\theta). } \end{align*} $$

and define the limit function by

$$ \begin{align*} \boxed{ g(\theta): = \sum_{n=1}^{\infty}\frac{n}{2^n}\cos(n\theta). } \end{align*} $$

Observe that

$$ \begin{align*} \left| \frac{n}{2^n}\cos(n\theta) \right| \leq \frac{n}{2^n} \end{align*} $$

for every \(\theta\). Hence set \(M_n = \frac{n}{2^n}\). We need to show that \(\sum_{n=1}^{\infty}M_n = \sum_{n=1}^{\infty}\frac{n}{2^n} < \infty\). To do this, let \(a_n=\frac{n}{2^n}\). Compute the ratio

$$ \begin{align*} \frac{a_{n+1}}{a_n} = \frac{\frac{n+1}{2^{n+1}}}{\frac{n}{2^n}} = \frac{n+1}{2n} = \frac{n}{2n}+\frac{1}{2n} = \frac12+\frac{1}{2n}. \end{align*} $$

Therefore,

$$ \begin{align*} \lim_{n\to\infty}\frac{a_{n+1}}{a_n} = \frac12<1. \end{align*} $$

By the ratio test, \(\sum_{n=1}^{\infty}\frac{n}{2^n}<\infty\). Therefore, by the Weierstrass \(M\)-test, the derivative series converges absolutely for every \(\theta\), and its sequence of partial sums \(S_N'\) converges uniformly to \(g\).

Now we have showed:

  1. \(S_N\to f\) uniformly.
  2. Each \(S_N\) is differentiable.
  3. \(S_N' \to g\) uniformly.

Hence by the term-by-term differentiation theorem,

$$ \begin{align*} \boxed{f'(\theta)=g(\theta).} \end{align*} $$

So \(f\) is indeed differentiable.

Finding the Fourier Coefficient

Since

$$ \begin{align*} f(\theta)=\sum_{n=1}^{\infty}\frac{1}{2^n}\sin(n\theta), \end{align*} $$

we substitute this into the definition:

$$ \begin{align*} \widehat f(k) = \frac{1}{2\pi} \int_{-\pi}^{\pi} \left( \sum_{n=1}^{\infty}\frac{1}{2^n}\sin(n\theta) \right)e^{-ik\theta}\,d\theta. \end{align*} $$

Because the series converges uniformly, we may interchange the infinite sum and the integral:

$$ \begin{align*} \widehat f(k) = \sum_{n=1}^{\infty} \frac{1}{2^n} \left[ \frac{1}{2\pi} \int_{-\pi}^{\pi} \sin(n\theta)e^{-ik\theta}\,d\theta \right]. \tag{1} \end{align*} $$

Let the integral be \(I_{n,k} = \frac{1}{2\pi} \int_{-\pi}^{\pi} \sin(n\theta)e^{-ik\theta} \,d\theta\). Using the identity \(\sin(n\theta)=\frac{e^{in\theta}-e^{-in\theta}}{2i}.\) to evaluate the \(I_{n,k}\):

$$ \begin{align*} I_{n,k} &= \frac{1}{2\pi} \int_{-\pi}^{\pi} \left( \frac{e^{in\theta}-e^{-in\theta}}{2i} \right)e^{-ik\theta}\,d\theta \\ &= \frac{1}{2i}\frac{1}{2\pi} \int_{-\pi}^{\pi} \left(e^{in\theta}-e^{-in\theta}\right) e^{-ik\theta}\,d\theta \\ &= \frac{1}{2i}\frac{1}{2\pi} \int_{-\pi}^{\pi} \left( e^{i(n-k)\theta} - e^{-i(n+k)\theta} \right)d\theta \\ &= \frac{1}{2i} \left[ \frac{1}{2\pi} \int_{-\pi}^{\pi}e^{i(n-k)\theta}\,d\theta - \frac{1}{2\pi} \int_{-\pi}^{\pi}e^{-i(n+k)\theta}\,d\theta \right]. \end{align*} $$

Recall the identity Use the orthogonality identity

$$ \begin{align*} \frac{1}{2\pi}\int_{-\pi}^{\pi}e^{im\theta}\,d\theta = \begin{cases} 1,&m=0,\\ 0,&m\neq 0. \end{cases} \end{align*} $$

Hence If \(k=n\), the first exponential becomes \(e^0=1\), while the second integral is zero. Hence

$$ \begin{align*} I_{n,n} = \frac{1}{2i}[1-0] = \frac{1}{2i}. \end{align*} $$

If \(k=-n\), the first integral is zero, while the second exponential becomes \(e^0=1\):

$$ \begin{align*} I_{n,-n} = \frac{1}{2i}[0-1] = -\frac{1}{2i}. \end{align*} $$

If \(k\neq n\) and \(k\neq -n\), both integrals are zero so \(I_{n,k} =0\). Hence

$$ \begin{align*} I_{n,k} = \frac{1}{2\pi} \int_{-\pi}^{\pi} \sin(n\theta)e^{-ik\theta}\,d\theta = \begin{cases} \dfrac{1}{2i},&k=n,\\[6pt] -\dfrac{1}{2i},&k=-n,\\[6pt] 0,&k\neq \pm n. \end{cases} \\ \tag{2} \end{align*} $$

Recall that

$$ \begin{align*} \widehat f(k) = \sum_{n=1}^{\infty}\frac{1}{2^n}I_{n,k}, \end{align*} $$

For a fixed \(k>0\), \(I_{n,k}\) is nonzero only when \(n = k\). Therefore, every term in the sum vanishes except the \(n\) term. Hence by (2)

$$ \begin{align*} \widehat f(k) = \frac{1}{2^k}\frac{1}{2i}. \end{align*} $$

When \(k < 0\), the only surviving term \(n = -k\). Since \(k<0\), the number \(-k\) is positive, as required because the original sum only uses \(n\geq1\). Therefore,

$$ \begin{align*} \widehat f(k) = \frac{1}{2^{-k}}\left(-\frac{1}{2i}\right) = -\frac{1}{2^{-k}\cdot 2i} \end{align*} $$

If we change the variable back to \(n\), we get

$$ \begin{align*} \widehat f(n)= \begin{cases} \dfrac{1}{2^n\cdot 2i},&n>0,\\[6pt] 0,&n=0,\\[6pt] -\dfrac{1}{2^{-n}\cdot 2i},&n<0. \end{cases} \end{align*} $$

References

  • SFSU Fourier Analysis by Professor Chun-Kit Lai
  • Fourier Analysis: An Introduction by Elias M.Stein and Rami Shakarchi