# Are Fourier transforms unique?

• pellman
In summary, the conversation discusses the relationship between two functions, g(x) and h(x), that have the same integral when multiplied by the exponential function, e^{ikx}. The question is whether this implies that g(x) = h(x) for all real values of x. The answer is yes, with the use of the Fourier inversion theorem and the concept of Lesbegue measure. By setting k = 0, it is shown that the two functions differ by a constant, which can be proven to be 0. However, this only applies to functions that have an inverse Fourier transform, and may not hold for all functions.
pellman
If

$$\int{g(x)e^{ikx}dx}=\int{h(x)e^{ikx}dx}$$

for all real k, can I conclude that g(x) = h(x) for all real x?

By setting k = 0 you know they differ by a constant. Show that this constant is 0.

pellman said:
If

$$\int{g(x)e^{ikx}dx}=\int{h(x)e^{ikx}dx}$$

for all real k, can I conclude that g(x) = h(x) for all real x?

You have to rely on Fourier inversion theorem, that you can find in any book that talks about Fourier transforms. In general, two functions (square integrable) that have the same Fourier transform are equal "almost everywhere", that is, everywhere except on a set of points with zero Lesbegue measure.

Jarle said:
By setting k = 0 you know they differ by a constant. Show that this constant is 0.

For k = 0 you only know they have the same average value, not that they differ by a constant.

Petr Mugver said:
For k = 0 you only know they have the same average value, not that they differ by a constant.

My apologies, you are right of course.

Assuming the functions are nice, so that the inverse Fourier transform exists, then g(x)=f(x) almost everywhere. The expression for the inverse transform looks just like the expression for the transform except for a sign reversal in the exponent.

In other words, let H(t) be the Fourier transform of f(x) or g(x), then the inverse transform of H(t) (call it h(x)) equals f(x) (and g(x)) almost everywhere.

Petr Mugver said:
In general, two functions (square integrable) that have the same Fourier transform are equal "almost everywhere", that is, everywhere except on a set of points with zero Lesbegue measure

Thanks! I will have to read up on Lesbegue measure.

## 1. What is a Fourier transform?

A Fourier transform is a mathematical tool used to decompose a mathematical function into its constituent frequencies. It transforms a function from its original time or spatial domain into a frequency domain.

## 2. Why are Fourier transforms important in science?

Fourier transforms are important in science because they allow us to analyze complex signals and systems in terms of their frequency components. This is useful in various fields such as signal processing, image processing, and quantum mechanics.

## 3. Are Fourier transforms unique?

Yes, Fourier transforms are unique. This means that a given function can only have one Fourier transform and vice versa. This is because the Fourier transform is a one-to-one mapping between a function and its frequency representation.

## 4. Can any function have a Fourier transform?

Technically, yes, any function can have a Fourier transform. However, the function must satisfy certain conditions such as being bounded, continuous, and having a finite number of discontinuities. Otherwise, the Fourier transform may not exist or may not accurately represent the function.

## 5. How do Fourier transforms relate to the Fourier series?

The Fourier transform is a generalized version of the Fourier series, which is used to represent periodic functions. While the Fourier series decomposes a function into a series of sine and cosine waves, the Fourier transform extends this concept to non-periodic functions by allowing for a continuous range of frequencies.

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