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
$$ h(x) = (f * g)(x) = \int_{\mathbb{R}^n } f(x-y) g(y) dy $$
Contents
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)
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 \(\widehat{F_c}(k)\): $$ \frac{d}{dk} \widehat{F_c}(k) = \frac{d}{dk} (\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \widehat{F_c}(x)e^{-ikx} dx) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \widehat{F_c}(x) \frac{d}{dk} e^{-ikx} dx = \frac{-i}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-cx^{2}}xe^{-ikx} dx $$ 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: $$ \frac{d}{dk} \widehat{F_c}(k) = -\frac{-i}{2c\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-cx^{2}}ike^{-ikx} dx $$
(not finsihed yet)
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: $$ \widehat{F_c}(0) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \widehat{F_c}(x) dx = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-cx^{2}} dx $$ 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 \(\phi(k) = \widehat{F_c}(k) = \frac{e^{-\frac{k^{2}}{4c}}}{\sqrt{2c}}\).
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: $$ e^{-ax^{2}}\sqrt{2\pi} \widehat{F_l}(s) = e^{-ax^{2}}\sqrt{2\pi} \frac{e^{-\frac{s^{2}}{4l}}}{\sqrt{2l}} = \sqrt{\frac{\pi}{l}} e^{-ax^{2}-\frac{4a^{2}x^{2}}{4l}} = \sqrt{\frac{\pi}{l}} e^{-(a+\frac{a^{2}}{l})x^{2}} = \sqrt{\frac{\pi}{a+b}} e^{-(a+\frac{a^{2}}{a+b})x^{2}} = α e^{-cx^{2}} = αF_c(x) $$ with \(α(a,b) = \sqrt{\frac{\pi}{a+b}}\) and \(c(a,b) = a+\frac{a^{2}}{a+b}\). The equation is valid since \(c = a+\frac{a^{2}}{a+b} > 0 + 0 = 0\).
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 \(f(x) = e^{-|x|}\), \(x \in \mathbb{R}\).