# Fourier transform of a unity function and of unit step function

Fourier transform of the unity function is the Dirac delta distribution.

I think this means:

In particular, the Fourier transform of the unity function is the Dirac delta distribution, $\mathcal F \mathbf 1 = \delta(x)$ and $\delta = u'$, % yleisesti ei oteta tassa reaaliarvoista funktiota where the step function is \begin{align} u(t) = \begin{cases} 0, & t < 0 \\ 1, & t \geq 0 \end{cases}. \end{align} The distributional derivative of the unit step function is the Dirac delta function $$\mathcal F u'(t) = \mathcal F \delta(t) = 1.$$

Is this correct?

• What do you mean by mark exactly ? Jan 10 '14 at 10:42
• @V.Rossetto I changed the question. I am not happy with this sentence The distributional derivative of the unit step function is the Dirac delta function. I know it can be said better to emphasize the Fourier transform. Jan 10 '14 at 10:44

This is actually a nasty business when one uses the Fourier transform of such things as a constant function on $\mathbf{R}$.

If you want to prove $u'=\delta$ in the distribution sense, you should use the correct way of calculating distribution derivatives as stated by L. Schwartz. Write $\left\langle f,\varphi\right\rangle$ the value of the distribution $f$ applied to a test function $\varphi\in\mathcal{C}^\infty$ with a compact support. You have $$\left\langle\delta,\varphi\right\rangle=\varphi(0).\tag 1$$ If the distribution $f$ is a locally integrable function then $$\left\langle f,\varphi\right\rangle=\int_{\mathbf{R}}f(x)\varphi(x)\,\mathrm{d}x.$$ Note that $\delta$ is not such a function, so the above formula does not apply for $\delta$.

According to your notations, one can compute $\left\langle u,\varphi\right\rangle$: $$\left\langle u,\varphi\right\rangle=\int_{\mathbf{R}}u(x)\varphi(x)\mathrm{d}x=\int_0^\infty\varphi(x)\mathrm{d}x.$$ To compute $u'$, apply now the definition of the derivative $f'$ $$\left\langle f',\varphi\right\rangle=-\left\langle f,\varphi'\right\rangle$$ to $f=u$. You get $$\left\langle u',\varphi\right\rangle=-\left\langle u,\varphi'\right\rangle=-\int_0^\infty\varphi'(x)\mathrm{d}x.$$ As $\varphi$ has a compact support, it is equal to $0$ at infinity and the above equation gives $$\left\langle u',\varphi\right\rangle=-\lim_{x\to\infty}\varphi(x)+\varphi(0)=\varphi(0)$$ which is exactly the definition of the distribution $\delta$ in equation (1). We just have proved that $u'=\delta$.

• You are right. Constant function is more accurate than unity function here. Can you prefer to some book or paper where you take this proof? Jan 10 '14 at 13:44

Fourier Transform of Unity Function is the Dirac delta distribution:

$\mathcal F${1}=$\delta(f)$

If you need the unit function, then the distributional derivative of the unit step function is the Dirac delta function:

$\mathcal F \{\tfrac{d u(t)}{dt}\} = \mathcal F \{\delta(t)\} = 1$

or

Fourier Transform of Unit Step Function:

$\mathcal F \{{u(t)}\} = \mathcal {\frac{1}{j2\pi f}} + \delta(f)$

where \begin{align} u(t) = \begin{cases} 0, & t < 0 \\ 1, & t \geq 0. \end{cases} \end{align}

• Do you think that this is the same as saying Fourier transform of the unity function is the Dirac delta distribution? Jan 9 '14 at 10:43
• What do you think if this is better written? $\mathcal F \mathbf 1 = \delta(x)$. Jan 9 '14 at 20:56
• Hmm, $\mathbf 1$ is identity. Is unity function the same as identity? Jan 9 '14 at 21:00