# 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 ? – Tom-Tom 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. – Léo Léopold Hertz 준영 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? – Léo Léopold Hertz 준영 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? – Léo Léopold Hertz 준영 Jan 9 '14 at 10:43
• @Masi, My bad! You are right, I read it backwards! – CAGT Jan 9 '14 at 17:02
• What do you think if this is better written? $\mathcal F \mathbf 1 = \delta(x)$. – Léo Léopold Hertz 준영 Jan 9 '14 at 20:56
• Hmm, $\mathbf 1$ is identity. Is unity function the same as identity? – Léo Léopold Hertz 준영 Jan 9 '14 at 21:00
• You want the step function: u(t) is 0 for t<0 and u(t)=1 for t>=0 ? Fourier transform I usually apply it from time domain to frequency domain, so I'm used to represent it that way. I'm not sure if there is any standard way of selecting which letter will be the independent variable in other applications. – CAGT Jan 9 '14 at 21:14