Continuity of Square Wave Fourier Series

Question

I am currently learning about Fourier series, and I am confused about a mismatch between two different results.

Let's take a square wave $\operatorname{x}\left(t\right)$ with period $1$ and amplitude $1$. The Fourier series is $$ \operatorname{x}\left(t\right) = \frac{4}{\pi}\sum_{n = 1}^{\infty} \frac{1}{2n - 1}\,\sin\left(2\pi\left[2n - 1\right]t\right) $$

• Each term in the summation is obviously continuous. Also, the sum of multiple continuous functions is continuous.
• Thus, I would expect the infinite sum to be continuous as well.
• However, the square wave clearly contains discontinuities, so I must be making a mistake somewhere. Is this a problem with infinity $?$. Is this some analysis concept I am unaware of $?$.

Thanks.

Answer 1

A sequence of continuous functions need not converge to a continuous function.

We can consider a simpler examples:

$$f_n(x)=x^n, x\in [0,1]$$

The limit is not a continuous function.

Forier-series allow us to construct many more such examples. Just take a piecewise continuous periodic function that is not continuous, the approximating polynomial is certainly continuous but the limit is not.

Answer 2

The function can be calculated explicitly. Namely let $$f(t)=\begin{cases} \phantom{-} 1 & \phantom{-} 0<t < \pi \\ -1 & -\pi <t<0 \end{cases}$$ Then the Fourier coefficient are $a_0=0$ and $$a_n={1\over \pi}\int\limits_{-\pi}^\pi f(t)\cos (nt)\,dt =0$$, $$ b_n={1\over \pi}\int\limits_{-\pi}^\pi f(t)\sin (nt)\,dt={2\over \pi}\int\limits_{0}^\pi\sin (nt)\,dt=\begin{cases}{4\over \pi(2k-1)} & n=2k-1\\ 0 & n=2k\end{cases} $$ Therefore, by the Dirichlet-Jordan condition we get $$f(t)={4\over \pi}\sum_{k=1}^\infty {\sin(2k-1)t\over 2k-1},\qquad 0<|t|<\pi$$ For the OP case we have $$x(t)=f(2\pi t)=\begin{cases}\phantom{-} 1 & \phantom{-} 0<t<{1\over 2} \\ -1 & -{1\over 2} <t<0\end{cases}$$ Therefore the function $x(t)$ is not continuous at $0,$ and after extension to $1$-periodic function is not continuous at ${n\over 2}.$

The reason for the continuity and discontinuity follows from the fact that the partial sums are uniformly convergent on each interval $(-{1\over 2}+\delta, -\delta)$ and $(\delta, {1\over 2}-\delta)$ but not uniformly convergent on $(-{1\over 2},0)$ and on $ (0,{1\over 2})$

Answer 3

"Each term in the summation is obviously continuous… Thus, I would expect the infinite sum to be continuous as well."

Err… why?

Properties that hold for all members of a sequence need not hold for the limit of the sequence (see Q757384). A natural example can be found with the truncated decimal expansions of $\sqrt{2}$. All of $$1,\,1.4,\,1.41,\,1.414,\ldots$$

are obviously rational. But the same cannot be said for $\sqrt{2}$, their limit.

To put it another way, being the limit of a sequence doesn't imply being similar to the sequence in other aspects. The only property a limit is actually concerned with is distance or size, such that $|x_n-L|$ can be made arbitrarily small for sufficiently large $n$. Hence, any other property not directly tied size can be lost in the limit (e.g., integer-ness or rationality for numbers, continuity or differentiability of functions, etc.).