For a function $f:\RR \to \CC$ we define:

$$\begin{align} \norm f_1 & = \int_{-\infty}^{\infty} \abs{f(x)} \dx \\ \norm f_2 & = \frac{1}{2\pi} \left(\int_{-\infty}^{\infty} \abs{f(x)}^2 dx\right)^{1/2} \\ \norm f_\infty & = \sup_x \abs{f(x)}. \end{align}$$

For $p=1,2,\infty$ we denote by

$$E^p(\RR) = \{f:\RR \to \CC \;\mid\; f \text{ piecewise continuous}, \norm f_p < \infty\}.$$

For each $p\in\{1,2,\infty\}$ find a function $f$ such that $\norm f_p < \infty$ , while the other two norms are infinity.

The map $\angles{f,g}$ defined above is an inner product on $E^2(\RR)$ .

Once we know that $\angles{f,g}$ is well defined, the rest of properties of inner products are easy to prove. We already saw in Inner product spaces - a reminder that the integral converges, but as a reminder, because $\abs{f(x)\overline{g(x)}}\leq \abs{f(x)}^2+\abs{g(x)}^2$ , we get that the integral even converges in absolute value since

$$\int_{-\infty}^\infty\abs{f(x)g(x)}\dx \leq \norm{f(x)}_2^2+\norm{g(x)}_2^2< \infty.$$

For a function $f:\RR \to \CC$ and $\omega \in \RR$ we write:

$$\hat f(\omega) := \frac{1}{2\pi} \int_{-\infty}^{\infty} f(x)e^{-i\omega x}dx = \angles{f,e^{i\omega x}}.$$

If this limit exists for every $\omega \in \RR$ , we call $\hat{f}:\RR \to \CC$ the Fourier transform of $f$ .

We sometimes denote this transform by $\hat{f} = \mathcal{F}[f]$ .

For $a< b$ define the characteristic function:

$$f(x)=\chi_{[a,b]}(x)=\cases{1&x\in[a,b]\\0&else}.$$

Computing its Fourier transform, we have:

$$2\pi \hat{f}(w)= \int_{-\infty}^{\infty} \chi_{[a,b]}(x)e^{-i\omega x}dx = \int_{a}^{b} e^{-i\omega x}dx =\frac{1}{\omega} ie^{-i\omega x}\mid_a^b = \frac{e^{-i\omega a}-e^{-i\omega b}}{\omega i}. $$

Note that the computation above is invalid when $\omega=0$ for which we have:

$$\hat{f}(0)=\frac{1}{2\pi} \int_{-\infty}^{\infty} \chi_{[a,b]}(x)dx=\frac {b-a}{2\pi}. $$

The function $\hat f(\omega)$ is easily seen to be continuous for $\omega \neq 0$ , but you can also check that it is continuous at $\omega = 0$ as well. Also, since $e^{i\omega x}$ is bounded, as $\omega \to \infty$ we see that $\hat{f}(\omega)\to 0$ . We will see later on that this is a much more general phenomena.

More over, when $a=-b$ is symmetric, the expression above is the real function:

$$\hat{f}(\omega)=\cases{\frac{\sin(\omega b)}{\omega \pi} & \omega \neq 0 \\ \frac{b}{\pi} & \omega = 0.}$$

First we note that $f(x):=e^{-a|x|}\in E^1(\RR)$ . Next, we have that

$$\begin{align} 2\pi \cdot \hat f(\omega) & = \int_{-\infty}^{\infty} e^{-a|x|} e^{-i\omega x}\dx = \int_{-\infty}^{0} e^{(a-i\omega) x}\dx + \int_{0}^{\infty} e^{(-a-i\omega) x}\dx \\ & = \int_{0}^{\infty}\left(e^{-(a-i\omega)x}+e^{(-a-i\omega)x}\right)\dx=\left[\frac{e^{-ax}e^{i\omega x}}{i\omega-a}-\frac{e^{-ax}e^{-i\omega x}}{i\omega+a}\right]_{0}^{\infty}. \end{align}$$

Since $e^{i\omega x}$ is bounded by 1, and $e^{-ax}\to 0$ as $x\to\infty$ , we conclude that

$$2\pi \cdot \hat f(\omega) = \frac{1}{i\omega+a}-\frac{1}{i\omega-a} = \frac{2a}{\omega^2+a^2}. $$

If $f\in E^1(\RR)$ , then $\hat{f}(\omega)$ is well defined for all $\omega$ and $\norm {\hat{f}}_\infty \leq \frac{1}{2\pi} \norm f_1$ .

This is a simple upper bound computation:

$$\abs{\hat{f}(\omega)}=\abs{\frac{1}{2\pi} \int_{-\infty}^{\infty} f(x)e^{-i\omega x}dx} \leq \frac{1}{2\pi} \int_{-\infty}^{\infty} \abs{f(x)}\abs{e^{-i\omega x}}dx=\frac{1}{2\pi} \norm f_1.$$

As we mentioned before, the “inner product” $\angles{f,e^{i\omega x}}$ is no longer a standard an inner product, however there is some method to this madness. This is is a product between a function in $L^1(\RR)$ and $L^\infty(\RR)$ , where it is usually between two functions in $L^2(\RR)$ . These two pairs of numbers satisfy the simple “equation” $\frac{1}{1}+\frac{1}{\infty} = \frac{1}{2} + \frac{1}{2} = 1$ , and as it turns out, this is enough for the “inner product” to be defined. For more details, you should read about Holder inequality.

Let $f\in E^1(\RR)$ . Then:

1. The Fourier transform is linear, namely $\mathcal{F}(\alpha f + g) = \alpha \mathcal{F}(f)+\mathcal{F}(g)$ .

2. Translation->Rotation: Setting $f_\alpha(x) := f(x+\alpha)$ , we have $$\hat{f}_\alpha (\omega) = e^{i\alpha\omega}\hat{f}(\omega).$$

3. Multiplication: If $\lambda \neq 0$ , then setting $g(x)=f(\lambda x)$ we have

   $$\hat g (\omega) = \frac {\hat{f}(\omega/\lambda)}{\abs{\lambda}}.$$

4. Conjugation: $\overline{\cf[f](\omega)}=\cf[\bar{f}](-\omega)$ .

1. Follows from the fact that integrals are linear.

2. (and 3) Both claims follow from change of parameters:

   $$\hat{f_\alpha}(\omega) = \frac{1}{2\pi}\int_{-\infty}^{\infty} f(x+\alpha)e^{-i\omega x}dx = \frac{1}{2\pi}\int_{-\infty}^{\infty} f(x)e^{-i\omega (x-\alpha)}dx = e^{i\omega\alpha}\hat f(\omega).$$

3. For $\lambda>0$ we have:

   $$\hat{g}(\omega) = \frac{1}{2\pi}\int_{-\infty}^{\infty} f(\lambda x)e^{-i\omega x}dx = \frac{1}{2\pi}\int_{-\infty}^{\infty} f(x)e^{-i(\omega/\lambda) x}\frac{1}{\lambda}dx = \frac{1}{\lambda}\hat f(\frac{\omega}{\lambda}).$$

   For $\lambda< 0$ , the proof is similar.

4. $$2\pi \overline{\hat{f}(\omega)} = \overline{\int_{-\infty}^{\infty} f(x) e^{-i\omega x}\dx} = \int_{-\infty}^{\infty} \overline{f(x)} e^{i\omega x}\dx = 2\pi \hat{\overline{f}}(-\omega)$$

Let $f\in E^1(\RR)$ and for $c\in \RR$ let $h_c(x):=e^{icx}f(x)$ . Then

$$\hat{h_c}(\omega)=\hat{f}(\omega-c).$$

In the sine and cosine notations, we have:

$$\align{ \cf[f(x)\sin(x)]&=\frac{\cf[f](\omega-c)-\cf[f](\omega+c)}{2i} \\ \cf[f(x)\cos(x)]&=\frac{\cf[f](\omega-c)+\cf[f](\omega+c)}{2} }$$

This is again a simple computation:

$$\hat{h_c}(\omega) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{icx}f(x)e^{-i\omega x}dx = \frac{1}{2\pi}\int_{-\infty}^{\infty} f(x)e^{-i(\omega-c)x}dx = \hat f(\omega-c).$$

If $f$ is even (resp. odd) then $\cf[f]$ is even (resp. odd).

If $f$ is real (resp. pure imaginary), then $\overline{\cf[f](\omega)}=\cf[f](-\omega)$ (resp. $=-\cf[f](-\omega)$ ).

In particular, if $f$ is real and even (e.g. cosine functions), then $\hat{f}$ is real and even, and if $f$ is real and odd, then $\hat{f}$ is pure imaginary and odd.

Take the multiplication identity with $\lambda = -1$ , so that $g(x)=f(-x)$ . In this case we get that $\hat{g}(\omega)=\hat{f}(-\omega)$ . If $f$ is symmetric, then $g=f$ , so that $\hat{f}(\omega)=\hat{f}(-\omega)$ is symmetric, and if $f$ is odd, then $g(x)=-f(x)$ in which case

$$\hat{f}(-\omega)=\widehat{g}(\omega)=\widehat{-f}(\omega)=-\widehat{f}(\omega)$$

is odd.

The second part is proved similarly from the conjugation identity.

If $f\in E^1(\RR)$ , then:

1. The function $\hat{f}$ is continuous.

2. (The Riemann-Lebesgue lemma): $\limfi{\abs{\omega}} \hat{f}(\omega) = 0$ .

1. For the first claim, let’s consider the following example with $f(x)=e^{-\abs{x}/10}$

   

   We would like to show that the integrals over $f(x)e^{-i\omega x}$ and $f(x) e^{-i\omega_0x}$ are close if $\omega, \omega_0$ are close. However, since it is hard to draw complex function, we will look at their real analogues. In the picture above, the function itself is drawn in green, and we also draw $f(x)\cdot \sin(10x)$ and $f(x)\cdot \sin(10.2 x)$ in blue and red. The transform should be thought of as the integrals over these function.

   In this example we want to show that since $10$ and $10.2$ are close, their integrals over $f(x)\cdot \sin(10x)$ and $f(x)\cdot \sin(10.2 x)$ are also close. There are two ideas that can be seen in the image that we can use in general:

   1. If we are “near the center” the blue and red functions themselves are very close so their integrals are close and

   2. If we are “far away from the center”, while this no longer holds, the total area beneath $f(x)$ is very small, so even when multiplying by some sine functions it remains small. Thus, even if we look at the difference between two such functions, the area will be small.

   More formally, fix some $M>0$ which “measures” how far we are from the center, then:

   $$\align{ 2\pi\abs{\hat{f}(\omega+h)-\hat{f}(\omega)} & \leq \int_{-\infty}^\infty\abs{e^{-i(\omega+h)x}-e^{-i\omega x}}\abs{f(x)}\dx = \int_{-\infty}^\infty\abs{e^{-ihx}-1}\abs{f(x)}\dx \\ & = \int_{\abs{x}\leq M}\abs{e^{-ihx}-1}\abs{f(x)}\dx + \int_{\abs{x}> M}\abs{e^{-ihx}-1}\abs{f(x)}\dx \\ & \leq \sup_{\abs{x}\leq M}\abs{e^{-ihx}-1} \cdot \int_{\abs{x}\leq M}\abs{f(x)}\dx + 2\int_{\abs{x}> M}\abs{f(x)}\dx \\ & \leq \sup_{\abs{x}\leq M}\abs{e^{-ihx}-1} \cdot \norm f_1 + 2\int_{\abs{x}> M}\abs{f(x)}\dx.}$$

   Choose your favorite $\varepsilon>0$ . Since $\norm f_1 < \infty$ , we can choose $M$ big enough so that

   $$2\int_{\abs{x}> M}\abs{f(x)}\dx < \frac {\varepsilon}{2}$$

   Next, for $\abs{x}\leq M$ , for all $\abs{h}$ small enough, using the continuity of $e^{-ihx}$ we can make sure that

   $$\sup_{\abs{x}\leq M}\abs{e^{-ihx}-1} \cdot \norm f_1< \frac{\varepsilon}{2}.$$

   Together we see that as $h\to 0$ we have that $\hat{f}(\omega+h)\to \hat{f}(\omega)$ , or in other words $\hat{f}$ is continuous.

2. The Riemann Lebesgue lemma is proved in a similar fashion.

   

   When computing the integral in $f(\omega)$ , when we are far away from the center the integral will be very small, regardless of $\omega$ . When we are close to the center, we can approximate our function by a step function. For each such step, when $\omega$ is very large, we should expect very high frequency fluctuations so every “positive area” will more or less cancel out with a “negative area”.

   More formally, for any choice of $M>0$ we get a simple upper bound:

   $$\align{ 2\pi\abs{\hat{f}(\omega)} & = \abs{\int_{-\infty}^\infty e^{-i\omega x}f(x)\dx} \leq \abs{\int_{\abs{x}\leq M} e^{-i\omega x}f(x)\dx} + \int_{\abs{x}>M} \abs{f(x)}\dx.}$$

   As before, for any $\varepsilon>0$ , we can choose $M$ big enough so that the second summand is as small as we want :

   $$\int_{\abs{x}>M} \abs{f(x)}\dx < \frac{\varepsilon}{2}.$$

   Since our function $f$ is Riemann integrable, we can approximate it by a step function $h$ , namely $h(x)=\sum_1^n \lambda_i \chi_{[a_i,b_i]}$ is a finite combination of steps, and $\int_{\abs{x}\leq M}\abs{h(x)-f(x)}\dx < \frac{\varepsilon}{4}$ . We can use it to upper bound the first summand:

   $$\align{ \abs{\int_{\abs{x}\leq M} e^{-i\omega x}f(x)\dx}&=\abs{\int_{\abs{x}\leq M} e^{-i\omega x}\left(f(x)-h(x)\right)\dx + \int_{\abs{x}\leq M} e^{-i\omega x}h(x)\dx} \\ & \leq \int_{\abs{x}\leq M} \abs{f(x)-h(x)}\dx + \abs{\sum_{i=1}^n\lambda_i \int_{\abs{x}\leq M} e^{-i\omega x}\chi_{[a_i,b_i]}(x)\dx} \\ & \leq \frac{\varepsilon}{4}+\sum_{i=1}^n\abs{\lambda_i}\hat{\chi}_{[a_i,b_i]}(\omega). }$$

   We already saw that $\hat{\chi}_{[a_i,b_i]}(\omega)=\frac{e^{-i\omega a}-e^{-i\omega b}}{\omega i}\to0$ as $\omega \to \infty$ , and since there are only finitely many such summands ( $n$ now is fixed), for all $\omega$ large enough this sum is at most $\frac{\varepsilon}{2}$ .

   Putting everything together, we get that for all $\varepsilon>0$ and for all $\omega$ large enough, we have that $\abs{\hat{f}(\omega)}< \varepsilon$ , or in other words $\hat{f}(\omega)\to 0$ .

Suppose that both $f\in E^1(\RR)$ , and $f'\in E^1 (\RR)$ . Then

$$\cf[f'](\omega)=i\omega \cf[f](\omega).$$

Similarly, if $xf(x)\in E^1 (\RR)$ , then $\hat{f}$ is differentiable and

$$\cf[-ixf(x)](\omega) = \cf[f]'(\omega).$$

Before we give a proper proof, note that if we are allowed to change the order of derivation and integration, then the second part is almost automatic. However, we don’t know that we can do this, and we basically prove it here.

For the first part, we use integration by parts

$$\align{ 2\pi \widehat{f'}(\omega)&=\int_{-\infty}^{\infty} f'(x)e^{-i\omega x}\dx=f(x)e^{-i\omega x}\mid_{-\infty}^{\infty}+i\omega\int_{-\infty}^{\infty} f(x)e^{-i\omega x}\dx. }$$

Using the fact that $f(x)=f(0)+\int_0^x f'(t)\dt$ and $\norm {f'}_1< \infty$ is finite, we see that $\limfi{x} f(x)$ exists. Since we also know that $\norm{f}_1< \infty$ is finite, this limit must be zero. Similarly, we get that $\displaystyle{\lim_{x\to -\infty}}f(x)=0$ , and since $e^{-i\omega x}$ is bounded, we conclude that

$$f(x)e^{-i\omega x}\mid_{-\infty}^{\infty} = 0. $$

The other summand is simply $2\pi i \omega \hat{f}(\omega)$ which completes the first part of claim.

For the second part, by definition we have that

$$\align{ 2\pi\left(\cf[f]'(\omega) -\cf[-ixf(x)] \right) & =\lim_{h\to 0}\int_{-\infty}^{\infty}f(x)\frac{e^{-i(\omega+h)x}-e^{-i\omega x}}{h}\dx + i\int_{-\infty}^{\infty}xf(x) e^{-i\omega x}\dx \\ & = \lim_{h\to 0}\int_{-\infty}^{\infty}f(x)e^{-i\omega x}\left( \frac{e^{-ihx}-(1-ihx)}{hx}x \right)\dx. }$$

As $e^z$ is analytic, we can find some constant $C$ such that $\abs{\frac{e^{-ihx}-(1-ihx)}{hx}}< C\abs{hx}^2$ for $\abs{hx}< 1$ . If $\abs{hx}>1$ , then $\abs{\frac{e^{-ihx}-(1-ihx)}{hx}}< 3$ . Using the fact that $\norm {xf(x)}_1 < \infty$ , the same ideas from before show that the limit is zero.

Note first that the function $f(x)$ is in $E^1(\RR)$ , so it has a Fourier transform.

For the $n=0$ case we know that $\cf\left[e^{-|x|}\right](\omega) = \frac{1}{\omega^2 + 1}$ . Using the last claim and a bit of induction, we get that :

$$\cf\left[x^ne^{-|x|}\right](\omega) = (-i)^n \cf\left[(-ix)^ne^{-|x|}\right](\omega) = (-i)^n \cf\left[e^{-|x|}\right]^{(n)}(\omega) = (-i)^n \frac{\partial^n}{\partial\omega^n}\left(\frac{1}{\omega^2+1}\right).$$

For example, for $n=1$ we simply get $\cf\left[xe^{-|x|}\right](\omega)=\frac{2\omega}{(\omega^2+1)^2}i$ .

Letting $f(x)=e^{-x^2}$ , its Fourier transform is

$$\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-x^2}e^{-i\omega x}\dx,$$

which doesn’t seem to simple to compute.

Instead, let us use the fact that we can take the derivative beneath the sign of the integral to get that :

$$ 2\pi \hat{f}'(\omega)= \int_{-\infty}^{\infty} e^{-x^2}(-ix)e^{-i\omega x}\dx = \frac{i}{2} \int_{-\infty}^{\infty} (-2x)e^{-x^2}e^{-i\omega x}\dx.$$

Taking $u(x)=e^{-x^2}$ , and $v(x)=e^{-i\omega x}$ , our integral is $u'(x)v(x)$ so we can do integration by parts. It follows that

$$ 2\pi \widehat{f}'(\omega)= \frac{i}{2} e^{-x^2}e^{-i\omega x}\mid_{-\infty}^{\infty} - \frac{1}{2}\omega \int_{-\infty}^{\infty} e^{-x^2}e^{-i\omega x}=\pi\omega \hat{f}(\omega).$$

The solution to the differential equation $\hat{f}'(\omega)=\frac{\omega}{2}\hat{f}(\omega)$ is $ae^{-\omega^2/4}$ . Finally, since

$$a = \hat{f}(0)=\frac {1}{2\pi} \int_{-\infty}^{\infty} e^{-x^2}\cdot 1 \dx = \frac{1}{2\sqrt{\pi}},$$

we conclude that

$$\hat{f}(\omega) = \frac{1}{2\sqrt{\pi}} e^{-\omega^2/4}.$$