Difference between revisions of "Aufgaben:Problem 4"

Revision as of 13:42, 27 December 2014

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 a)

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:

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

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

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