Aufgaben:Problem 4

1) Let \( f,g:\mathbb{R}^n \rightarrow \mathbb{C} \) be continuous, bounded. Further, \( f,g \in L^1(\mathbb{R}^n) \). Define the convolution \(h:\mathbb{R}^n \rightarrow \mathbb{C} \) as

$$ h(x) = (f * g)(x) = \int_{\mathbb{R}^n } f(x-y) g(y) dy $$

Part a)

Show that:

$$ (f*g)(x) = (g*f)(x) $$

Solution part a)

$$ (f*g)(x) = \int_{\mathbb{R}^n } f(x-y) g(y) dy $$

$$ = \int_{-\infty}^{\infty} ... \int_{-\infty}^{\infty} f(x-y) g(y) dy_1 ... dy_n $$

The claim follows simply by substitution:

$$ t = x-y $$

$$ y = x-t $$

$$ dy_i = -dt_i $$

$$ \int_{\infty}^{-\infty} ... \int_{\infty}^{-\infty} f(t) g(x-t) (-dt_1) ... (-dt_n) $$

We can now change the orientation of the integral if we multiply (-1) n-times. These will cancel out all the n minus-signs from the \( dt_i\)`s. Thus we get:

$$ \int_{-\infty}^{\infty} ... \int_{-\infty}^{\infty} f(t) g(x-t) dt_1 ... dt_n $$

$$ = \int_{\mathbb{R}^n } f(t) g(x-t) dt = (g*f)(x) $$

This proves the claim. \( \square \)

Part b)

Show that:

$$ \widehat{f*g}(k) = (2\pi)^{n/2} \hat f (k) \hat g (k) $$

Solution part b)

The definition of the fourier coefficient is:

$$ \widehat{f*g}(k) = \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n } (f*g)(x) e^{-i<k,x>} dx $$

And with the defintion of the convolution:

$$ = \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n } \int_{\mathbb{R}^n } f(x-y) g(y) dy \; e^{-i<k,x>} dx $$

Because \( e^{-i<k,x>} \) is independent of y, we can put it inside the inner integral.

$$ = \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n } \int_{\mathbb{R}^n } f(x-y) g(y) e^{-i<k,x>} dy \: dx $$

Now we can make a similar substitution as in part a)

$$ x = t+y $$

$$ t = x-y $$

$$ dx = dt $$

$$ = \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n } \int_{\mathbb{R}^n } f(t) g(y) e^{-i<k,t+y>} dy \: dt $$

With the bilinearity of the scalar product we can write \(<k,t+y>\) as \(<k,t> + <k,y>\). We then get

$$ \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n } \int_{\mathbb{R}^n } f(t) g(y) e^{-i<k,t>-i<k,y>} dy \: dt $$

$$ = \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n } \int_{\mathbb{R}^n } f(t) e^{-i<k,t>} g(y) e^{-i<k,y>} dy \: dt $$

As we can see, \( f(t) e^{-i<k,t>} \) is independent of y and \(g(y) e^{-i<k,y>}\) is independent of t. Therefore we can write it as two seperate integrals:

$$ = \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n } f(t) e^{-i<k,t>} dt \: \int_{\mathbb{R}^n } g(y) e^{-i<k,y>} dy $$

$$ = (2\pi)^{n/2} \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n } f(t) e^{-i<k,t>} dt \: \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n } g(y) e^{-i<k,y>} dy $$

$$ = (2\pi)^{n/2} \hat f (k) \hat g (k) \; \square $$

Part c)

Let \( 0 < c < \infty<\), and define \( F_c(x) = e^{-cx^{2}} \).

Compute \(\widehat{F_c}(k)\) by first showing that \(2c\phi'(k) + k\phi(k) = 0\), with \(\phi = \widehat{F_c}\), and then solving the differential equation.

Solution

We proof that \(2c\widehat{F_c}'(k) + k\widehat{F_c}(k) = 0\). First we try to bring \(\widehat{F_c}'(k)\) into a form that is related to \(\widehat{F_c}(k)\): $$ \frac{d}{dk} \widehat{F_c}(k) = \frac{d}{dk} (\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} F_c(x)e^{-ikx} dx) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} F_c(x) \frac{d}{dk} e^{-ikx} dx = \frac{-i}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-cx^{2}}xe^{-ikx} dx $$ Now we use partial integration to rearrange the integral. We integrate \(xe^{-cx^{2}}\) and derive \(e^{-ikx}\). We get: $$ \frac{-i}{\sqrt{2\pi}} ([-\frac{e^{-cx^{2}}}{2c}e^{-ikx}]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \frac{e^{-cx^{2}}}{2c}ike^{-ikx} dx) $$ The first term vanishes. It follows: $$ \frac{d}{dk} \widehat{F_c}(k) = -\frac{k}{2c\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-cx^{2}}e^{-ikx} dx = -\frac{k}{2c} \widehat{F_c}(k) $$ This gives us: $$ 2c\widehat{F_c}'(k) + k\widehat{F_c}(k) = -k\widehat{F_c}(k) + k\widehat{F_c}(k) = 0 $$

Now we solve the differential equation \(2c\phi'(k) + k\phi(k) = 0\) by seperation of variables: $$ 2c\phi'(k) + k\phi(k) = 0 $$ $$ \phi' = -\frac{k}{2c}\phi $$ $$ \frac{d\phi}{dk} = -\frac{k}{2c}\phi $$ $$ \int \frac{d\phi}{\phi} = -\frac{1}{2c} \int k dk $$ $$ ln(\phi) = -\frac{k^{2}}{4c} + p $$ $$ e^{ln(\phi)} = e^{-\frac{k^{2}}{4c} + p} $$ $$ \phi = re^{-\frac{k^{2}}{4c}} $$ while \(p\) and \(r\) are integration constants and \(r = e^{p}\). To evaluate the constant \(r\) we first create a boundary condition: $$ \widehat{F_c}(0) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} F_c(x) dx = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-cx^{2}} dx $$ With the substitution \(x = \frac{t}{\sqrt{c}}\) we get \(dx = \frac{dt}{\sqrt{c}}\) and thus: $$ \widehat{F_c}(0) = \frac{1}{\sqrt{2\pi c}} \int_{-\infty}^{\infty} e^{-t^{2}} dt = \frac{1}{\sqrt{2\pi c}} \sqrt{\pi} = \frac{1}{\sqrt{2c}} $$ using the Gaussian Integral. It follows: $$ \phi(0) = \widehat{F_c}(0) $$ $$ r = \frac{1}{\sqrt{2c}} $$ and thus \(\phi(k) = \widehat{F_c}(k) = \frac{1}{\sqrt{2c}}e^{-\frac{k^{2}}{4c}}\).

Part d)

Show that \(F_a*F_b = αF_c\), where \(c = c(a,b)\) and \(α = α(a,b)\).

Solution

$$ (F_a*F_b)(x) = \int_{-\infty}^{\infty} F_a(x-y)F_b(y) dy = \int_{-\infty}^{\infty} e^{-a(x-y)^{2}}e^{-by^{2}} dy = \int_{-\infty}^{\infty} e^{-a(x^{2}-2xy+y^{2})}e^{-by^{2}} dy = e^{-ax^{2}} \int_{-\infty}^{\infty} e^{-(a+b)y^{2}}e^{2axy} dy $$

$$ = e^{-ax^{2}} \int_{-\infty}^{\infty} e^{-(a+b)y^{2}}e^{-i(\frac{2ax}{-i})y} dy = e^{-ax^{2}} \frac{\sqrt{2\pi}}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-ly^{2}}e^{-isy} dy = e^{-ax^{2}}\sqrt{2\pi} \widehat{F_l}(s) $$

with \(l = a + b\), \(s = -\frac{2ax}{i}\). The equation is valid since \(l = a + b > 0 + 0 = 0\). Using the identity evaluated in 1.c) we get:

$$ e^{-ax^{2}}\sqrt{2\pi} \widehat{F_l}(s) = e^{-ax^{2}}\sqrt{2\pi} \frac{1}{\sqrt{2l}}e^{-\frac{s^{2}}{4l}} = \sqrt{\frac{\pi}{l}} e^{-ax^{2}+\frac{4a^{2}x^{2}}{4l}} = \sqrt{\frac{\pi}{l}} e^{-(a-\frac{a^{2}}{l})x^{2}} = \sqrt{\frac{\pi}{a+b}} e^{-(a-\frac{a^{2}}{a+b})x^{2}} = α e^{-cx^{2}} = αF_c(x) $$

with \(α(a,b) = \sqrt{\frac{\pi}{a+b}}\) and \(c(a,b) = a-\frac{a^{2}}{a+b}\). The equation is valid since \(c = a-\frac{a^{2}}{a+b} > 0 \). This is because \(\frac{a^{2}}{a+b} < a \) for \(a, b > 0\).

Part e)

Let \(g \in L^{1}(\mathbb{R})\), \(u \in L^{1}(\mathbb{R}) \cup C^{2}(\mathbb{R})\). Find a solution of the differential equation $$ u^{(2)}(x) = u(x) - 2g(x) $$ using Fourier transform. Hint: consider the Fourier transform of \(f(x) = e^{-|x|}\), \(x \in \mathbb{R}\).

Solution

We first calculate the Fourier transform of \(f(x) = e^{-|x|}\), \(x \in \mathbb{R}\): $$ \widehat{f}(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-|x|}e^{-ikx} dx = \frac{1}{\sqrt{2\pi}} (\int_{-\infty}^{0} e^{-|x|}e^{-ikx} dx + \int_{0}^{\infty} e^{-|x|}e^{-ikx} dx) = \frac{1}{\sqrt{2\pi}} (\int_{-\infty}^{0} e^{x}e^{-ikx} dx + \int_{0}^{\infty} e^{-x}e^{-ikx} dx) $$ $$ = \frac{1}{\sqrt{2\pi}} (\int_{-\infty}^{0} e^{x(1-ik)} dx + \int_{0}^{\infty} e^{-x(1+ik)} dx) = \frac{1}{\sqrt{2\pi}} ([\frac{e^{x(1-ik)}}{1-ik}]_{-\infty}^{0} + [\frac{e^{-x(1+ik)}}{-(1+ik)}]_{0}^{\infty}) = \frac{1}{\sqrt{2\pi}} (\frac{1}{1-ik} - \frac{1}{-(1+ik)}) $$ $$ = \frac{1}{\sqrt{2\pi}} \frac{1+ik+1-ik}{1-i^{2}k^{2}} = \frac{1}{\sqrt{2\pi}} \frac{2}{1+k^{2}} $$ Now to the differential equation: $$ u^{(2)}(x) = u(x) - 2g(x) $$ $$ \widehat{u^{(2)}}(k) = \widehat{(u - 2g)}(k) $$ $$ \widehat{u^{(2)}}(k) = \widehat{u}(k) - 2\widehat{g}(k) $$ $$ i^{2}k^{2}\widehat{u}(k) = \widehat{u}(k) - 2\widehat{g}(k) $$ using \(\widehat{(\lambda\psi + \mu\varphi)} = \lambda\widehat{\psi} + \mu\widehat{\varphi}\) and \(\widehat{(\frac{\partial}{\partial x_j}\psi)}(k) = ik_j\widehat{\psi}(k)\) for \(\psi,\varphi \in L^{1}(\mathbb{R})\) and \(\lambda,\mu \in \mathbb{C}\). $$ -k^{2}\widehat{u}(k) = \widehat{u}(k) - 2\widehat{g}(k) $$ $$ \widehat{u}(k) (1 + k^{2}) = 2\widehat{g}(k) $$ $$ \widehat{u}(k) = \frac{2}{1+k^{2}}\widehat{g}(k) = \frac{\sqrt{2\pi}}{\sqrt{2\pi}} \frac{2}{1+k^{2}}\widehat{g}(k) = \sqrt{2\pi}\widehat{f}(k)\widehat{g}(k) $$ using the identity for \(\widehat{f}(k)\) evaluated in the first part of the exercise. Applying the inverse Fourier transform we get: $$ u(x) = \sqrt{2\pi} (\widehat{f}(k)\widehat{g}(k))^{\vee}(x) = \frac{\sqrt{2\pi}}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \widehat{f}(k)\widehat{g}(k)e^{ikx} dk = \int_{-\infty}^{\infty} \widehat{f}(k)\widehat{g}(k)e^{ikx} dk $$