TheFourierTransformOfZero

Theorem:

The Fourier transform of the constant zero function $f(x) = 0$ is $\delta(y)$, the Dirac delta function.

Proof:

The Fourier Transform $F(y)$ of a function $f(x)$ is defined as:

$$F(y) = \int_{-\infty}^{\infty} f(x) e^{-i 2 \pi y x} dx$$

When $f(x) = 0$, the integral becomes:

$$F(y) = \int_{-\infty}^{\infty} 0 \cdot e^{-i 2 \pi y x} dx = 0$$

However, this is an improper integral in the sense that the function we're integrating isn't a traditional function but rather a distribution.

When dealing with distributions, it's often more useful to consider how they act on test functions $\phi(y)$ rather than their values at particular points. In this case, the Dirac delta distribution $\delta(y)$ is defined such that for any smooth function $\phi(y)$ with compact support $$\int_{-\infty}^{\infty} \delta(y) \phi(y) dy = \phi(0)$$

Let $F(y) = \delta(y)$. Then for any $\phi(y)$ $$\int_{-\infty}^{\infty} F(y) \phi(y) dy = \int_{-\infty}^{\infty} \delta(y) \phi(y) dy = \phi(0)$$

which is exactly the same as integrating $f(x) = 0$ against any test function $\phi(x)$:

$$\int_{-\infty}^{\infty} 0 \cdot \phi(x) dx = 0 = \phi(0)$$

Therefore, in the sense of distributions, the Fourier transform of $f(x) = 0$ is $\delta(y)$.

End of Proof.