Fourier transform of function defined on finite interval

Let $$f(t)$$ be a function defined on the finite interval $$[t_1,\, t_2]$$. Is the Fourier transform of such a function uniquely defined? In the sense that there exists only one function $$\hat{f}(\omega)$$ giving the Fourier coefficients that sum up to $$f(t)$$ in the interval $$[t_1,\,t_2]$$?

I am aware that one can find the Fourier transform of a function $$g_1(t)$$ defined on $$[-\infty, \,+\infty]$$ as
$$g_1(t) = \begin{cases} f(t) & \text{for}\;t \in [t_1, t_2]\\ 0 & \text{for}\;t \lt t_1 \;\text{and}\;t \gt t_2 \end{cases}$$

The Fourier transform of this function would yield the coefficients that sum up to $$f(t)$$ on the interval $$[t_1,\,t_2]$$, I think. However, the same could be said about functions $$g_2(t)$$ and $$g_3(t)$$ defined via $$g_2(t) = \begin{cases} f(t) & \text{for}\;t \in [t_1, t_2]\\ 4 & \text{for}\;t \lt t_1 \;\text{and}\;t \gt t_2 \end{cases}$$ $$g_3(t) = \begin{cases} f(t) & \text{for}\;t \in [t_1, t_2]\\ t & \text{for}\;t \lt t_1 \;\text{and}\;t \gt t_2 \end{cases}$$ or any other function that matches the definition of $$f(t)$$ on $$[t_1,\,t_2]$$ and somehow continues this to be defined on $$[-\infty,\,+\infty]$$. In particular, by continuing $$f(t)$$ periodically on $$[-\infty,\,+\infty]$$, one would get a Fourier series that would also be a valid representation of $$f(t)$$ on $$[t_1,\,t_2]$$. Or is this reasoning incorrect? If not, in what way does the concept of a Fourier transform apply to functions that are only defined on a finite interval?

• It doesn't make sense as such. The transform applies to functions defined on $\mathbb{R}$. – copper.hat Jul 9 '19 at 22:52
• – Maximilian Janisch Jul 9 '19 at 23:05

In mathematics, there are many Fourier transforms. For every locally compact abelian group, there is a corresponding Fourier transform. Two of these Fourier transforms can be applied to a function $$f:[t_1,t_2]\rightarrow\mathbb{C}.$$ 1) You can apply the Fourier transform $$\mathscr{F}_{\mathbb{T}}$$, where $$\mathbb{T}:=\mathbb{R}/(t_2-t_1)\mathbb{Z}$$ is a torus group. An application of $$\mathscr{F}_{\mathbb{T}}$$ corresponds to the action of extending $$f$$ periodically and then expanding it into a Fourier series. If you choose $$\mathscr{F}_{\mathbb{T}}$$ as your Fourier transform, $$\mathscr{F}_{\mathbb{T}}(f)$$ will be unique.

2) You can also choose to apply the Fourier transform $$\mathscr{F}_{\mathbb{R}}$$, but then you have to extend $$f$$ from $$[t_1,t_2]$$ to $$\mathbb{R}$$ beforehand. As you pointed out yourself, there is no unique way of doing this. Consequently, there is no unique way to define $$\mathscr{F}_{\mathbb{R}}(f)$$.

If you consider the finite abelian group $$\mathbb{Z}_N:=\mathbb{Z}/N\mathbb{Z}$$, the corresponding Fourier transform $$\mathscr{F}_{\mathbb{Z}_N}$$ is called distrete Fourier Transform. If you only have $$N$$ sampling points $$f(x_0),f(x_1),\ldots,f(x_{N-1})$$, you may think of it as a function on $$\mathbb{Z}_N$$ and consequently apply the Fourier transform $$\mathscr{F}_{\mathbb{Z}_N}$$ to it.

I will not define the Fourier transforms $$\mathscr{F}_{\mathbb{T}}$$, $$\mathscr{F}_{\mathbb{R}}$$, $$\mathscr{F}_{\mathbb{Z}_N}$$ here (you can easily find good sources elsewhere), but simply emphasize that there is more than one Fourier transform, which may resolve some of the perplexity behind the question in the OP.

• So to sum up, if a function is only given on a finite interval, one has to extend it over $\mathbb{R}$ in one way or another in order to be able to compute an expansion in $e^{-i \omega t}$ basis functions (resulting in either a Fourier transform or series). – Quantum Jul 30 '19 at 18:34
• The reason I am asking this question is that for the discrete Fourier transform, the samples $f[t_n]$ are only given on a finite interval. So I thought that finding the Fourier transform for the function represented by the samples is a bit arbitrary unless some assumptions are made about what the function does outside that interval. Any comments on this? – Quantum Jul 30 '19 at 18:37
• @Quantum: The discrete Fourier transform is yet another Fourier transform. If you only have a finite number of sampling points, you can apply it. You don't need to interpolate in order to apply it. – StarBug Jul 30 '19 at 20:23
• Sure, I do not need to interpolate within the interval. But does the discrete transform not make any assumption (even if only implicitly) about the function outside the sampled interval (such as being zero or periodic)? If the discrete transform is considered an approximation to the continuous transform or series, then it seems to me that some extension has to be made outside the interval for the reasons you described above. – Quantum Jul 31 '19 at 8:35
• That clarifies it, thank you, although I have not seen this stated explicitly in treatments of the DFT. I guess what I really meant with my original question was "for a function defined on a finite interval, is the expansion in sinusoids/complex exponentials unique?". It is now clear that the answer to that is "no, there are several ways of expanding it, some of which assume an extension over the real numbers". – Quantum Jul 31 '19 at 21:11