Why is a Fourier series expansion not always equal to the function being expanded?

Question

It is generally taught in calculus classes that it is not always true that the Fourier series expansion is equal to the function being expanded into Fourier series. The series can actually fail to converge entirely, and if it does converge, it could converges to something totally different than the function itself.

Over 15+ years experience in the topic, I have never encountered such a situation so far. I am curious is to see at least one example or two highlighting this interesting fact. Thank you!

You can take say $f=1$ for $x\in[0,\pi]$ and $f=0$ for $x\in(-\pi,0)$, then $f(0)=1$ but the Fourier series iirc will take the value $1/2$ at $x=0$. I would guess, a milder failure than what is possible — Calvin Khor, Apr 28 '23 at 07:35
@CalvinKhor I see. So this happens only at interval discontinuities. Just curious if continuous functions can show such a behavior... — keynes, Apr 28 '23 at 07:37
A continuous map can have a divergent Fourier series. See here for an example. The divergence can also occur at a dense subset. — mathcounterexamples.net, Apr 28 '23 at 07:42
it took a while for me to jog the memory, but there is the following consequence of the uniform boundedness principle: there is a continuous function whose Fourier series diverges at 0. See e.g. https://math.stackexchange.com/questions/3406915/what-is-the-baire-category-argument-here-divergence-of-many-fourier-series-at — Calvin Khor, Apr 28 '23 at 07:42
@mathcounterexamples.net Merci Jean-Pierre. So this could be seen in functions defined as infinite sums. Interesting. Greetings to Paris. — keynes, Apr 28 '23 at 07:50
And Wikipedia convergence of Fourier series provides conditions for convergence as well as example of divergence. — mathcounterexamples.net, Apr 28 '23 at 07:50
@mathcounterexamples.net You mentioned on your homepage that it is possible to switch between the int and sum sings if the series is n o r m a l l y convergent. Isn't u n i f o r m convergence enough to do that? — keynes, Apr 28 '23 at 08:36
@Staufenberg Yes uniform convergence is enough. But normally convergence implies uniform convergence, and is often simple to prove. — mathcounterexamples.net, Apr 28 '23 at 11:15

acat3 · Accepted Answer · 2023-04-28T08:22:16.410

Orthonormal Expansion

To properly discuss about Fourier Series, we have to understand orthonormal basis expansion / approximation. Briefly, a point in a vector space can be approximated by linear combination of basis for another space:

$$ \mathbf{w} \approx \tilde{\mathbf{w}} = \sum_{m=1}^{d} w_{m} \mathbf{v}_{m} , \phantom{x} \mathbf{v}_{m}\in\mathbb{V} \phantom{x} \forall \phantom{x}m \in \left\{1,...,d\right\} $$

We can define a criteria to evaluate how well this approximation perform e.g. we define the criteria as $L_{2}$ error:

$$ e= \left\| \mathbf{w} - \tilde{\mathbf{w}}\right\|^{2} = \left\langle \mathbf{w} - \tilde{\mathbf{w}}, \mathbf{w} - \tilde{\mathbf{w}} \right\rangle $$

If the basis are orthonormal, the scalars $w_{i}$ that minimises the aforementioned error is given by the following expression:

$$ w_{m} = \left\langle\mathbf{v}_{m}, \mathbf{w}\right\rangle $$

Worth noting that the approximation $\tilde{\mathbf{w}}$ is the projection of $\mathbf{w}$ to $\mathbb{V}$.

Connecting The Dots

Fourier series expansion is an orthonormal expansion. It approximate a function using some orthonormal basis for sinusoidal function space:

$$ f(t) \approx \tilde{f}(t) = \sum_{m=1}^{d} f_{m}\cdot \frac{1}{\sqrt{T}} e^{j2\pi m \frac{t}{T}} $$

The inner product is defined as the following:

$$ \left\langle f(t),g(t) \right\rangle = \int_{0}^{T} \bar{f} (t)\cdot g(t) \phantom{.}dt $$

The Fourier series coefficient is given by:

$$ f_{i} = \left\langle \frac{1}{\sqrt{T}}e^{j2\pi m \frac{t}{T}}, f(t) \right\rangle = \frac{1}{\sqrt{T}}\int_{0}^{T} e^{-j2\pi m \frac{t}{T}} \cdot f(t) \phantom{.}dt $$

I hope that last part looks familiar :)

Remarks

We have seen that Fourier series is an approximation of a function using sinusoidal function. Most of the time, approximation is not equal to the function being approximated. Some guiding questions:

If you approximate a discontinuous function using some continuous functions, can the approximation be exactly equal to the function being approximated?
If you approximate a non-periodic function using some periodic functions (with the same period), can the approximation be exactly equal to the function being approximated?
If you approximate a function that is not infinitely continuously differentiable using some functions functions that are, can the approximation be exactly equal to the function being approximated?

In many engineering / science program, Fourier transform & Fourier series are taught without solid basics of functional analysis and linear algebra. Therefore, students often fail to see them as what they are: approximation and projection.

The answer to your question If you approximate a function that is not infinitely continuously differentiable using some functions that are, can the approximation be exactly equal to the function being approximated? is yes. For example, if the initial function is $\mathcal C^1$ but not twice differentiable. — mathcounterexamples.net, Apr 28 '23 at 15:37

Why is a Fourier series expansion not always equal to the function being expanded?

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