Aufgaben:Problem 12

Part a)

Let \( \Phi_0(x) = e^{-\frac{1}{2}x^2} \) and define \( \Phi_n(x) = (-1)^n e^{\frac{1}{2}x^2}\left(\left(\frac{d}{dx}\right)^n e^{-x^2}\right) \). Show that \( \hat \Phi_n(k) = C\Phi_n(k)\), with \( C = C(n) \in \mathbb{C} \).

Solution

We show this by proving the following claim:

\( \vdash : \Phi_n(x) = (-1)^n e^{\frac{1}{2}x^2} \Big(\frac{d}{dx}\Big)^n e^{-x^2}\) is an eigenfunction of the Fourier-transform with the eigenvalue \((-i)^n \).

Proof.

We first check the case \( \hat \Phi_0(k) \) where we complete the square in the exponent by using the substitution \( \eta \equiv \frac{x + ik}{\sqrt{2}} \):

$$ \hat \Phi_0(k) = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R} } e^{-\frac{1}{2}x^2} e^{-ikx} dx = \frac{\sqrt{2}}{\sqrt{2\pi}} \int_{\mathbb{R} } e^{-\eta^2 - \frac{k^2}{2}} d\eta = e^{-\frac{1}{2} k^2} = \Phi_0(k) \tag{3.14} $$

For the general case:

$$ \sqrt{2\pi} \hat \Phi_n(k) = \int_{\mathbb{R} } (-1)^n e^{\frac{1}{2}x^2} \Big( (\frac{d}{dx})^n e^{-x^2} \Big) e^{-ikx} dx = \int_{\mathbb{R} } e^{-x^2} \Big( \frac{d}{dx} \Big)^n e^{\frac{1}{2}x^2 - ikx} dx $$

where we partially integrated. The first term of each partial integration is zero since:

$$ \left| \left( \frac{d}{dx} \right)^m e^{\frac{1}{2}x^2 - ikx} \left( \left(\frac{d}{dx}\right)^l e^{-x^2} \right) \right| = \left| p(x) \right| e^{-\frac{1}{2}x^2} $$

where \( p(x) = \mathcal{O}(n^j) \) (using the Landau notation) and \(l, m, j \in \mathbb{N} \) leading to

$$ \left| p(x) \right| e^{-\frac{1}{2}x^2} \bigg \vert_{-\infty}^{\infty} = 0 $$

Furthermore

$$ \sqrt{2\pi} \Phi_n(k) = e^{\frac{1}{2}k^2} \int_{\mathbb{R} } e^{-x^2} \Big( \frac{d}{dx} \Big)^n e^{\frac{1}{2}(x - ik)^2} dx = i^n e^{\frac{1}{2}k^2} \int_{\mathbb{R} } e^{-x^2} \Big( \frac{d}{dk} \Big)^n e^{\frac{1}{2}(x - ik)^2} dx $$

where we used the following

Lemma 1: \( \forall n \in \mathbb{Z}_{\geqslant 0}: i^n \Big( \frac{d}{dk} \Big)^n e^{\frac{1}{2}(x - ik)^2} = \Big( \frac{d}{dx} \Big)^n e^{\frac{1}{2}(x - ik)^2} \)

Proof of Lemma 1:

By induction.

\( n = 0 \) : \( i^0 e^{\frac{1}{2}(x - ik)^2} = e^{\frac{1}{2}(x - ik)^2} \) is the trivial case.

\( n-1 \rightarrow n \) :

$$ i^n \Big( \frac{d}{dk} \Big)^n e^{\frac{1}{2}(x - ik)^2} = i^{n-1} \Big( \frac{d}{dk} \Big)^{n-1} i (-i) (x-ik) e^{\frac{1}{2}(x - ik)^2} = i^{n-1} \Big( \frac{d}{dk} \Big)^{n-1} (x-ik) e^{\frac{1}{2}(x - ik)^2} = i^{n-1} \Big( \frac{d}{dk} \Big)^{n - 1} \frac{d}{dx} e^{\frac{1}{2}(x - ik)^2} = i^{n-1} \frac{d}{dx} \Big( \frac{d}{dk} \Big)^{n - 1} e^{\frac{1}{2}(x - ik)^2} $$

where for the last equality we used the Schwarz's theorem. With the induction assumption everything follows. \( \square \)

Lemma 2: \( \frac{d}{dk} \hat f(k) = \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} \frac{d}{dk} f(x) e^{-ikx} dx \) where \( f(x) = e^{- \frac{1}{2} x^2 - ikx} \).

Proof of Lemma 2:

$$ \sqrt{2\pi} \frac{d}{dk} \hat f(k) = \lim_{h \rightarrow 0} \int_{\mathbb{R}} \frac{1}{h} e^{- \frac{x^2}{2} - ikx} \left( e^{-ixh} - 1 \right) dx $$

We use \( \vert \sin(x) \vert = \vert x - \frac{x^3}{6} + \frac{x^5}{120} \pm\cdots \vert \leq \vert x \vert \) for

$$ \begin{align} \left| \frac{1}{h} e^{- \frac{x^2}{2} - ikx} \left( e^{-ixh} - 1 \right) \right| &= e^{- \frac{x^2}{2}} \frac{1}{h} \left| e^{-ihx} e^{-\frac{ixh}{2}} \left( e^{- \frac{ixh}{2} } - e^{\frac{ixh}{2}} \right) \right| \\ &= e^{- \frac{x^2}{2}} \frac{2}{h} \left| \sin \left( \frac{xh}{2} \right) \right| \leq \frac{2}{h} \left| \frac{xh}{2} \right| e^{-\frac{x^2}{2}} \\ &= \vert x \vert e^{-\frac{x^2}{2}} \end{align} $$

and with the Lebesgue dominated convergence theorem the claim holds. \( \square \)

The rest is a piece of cake:

$$ \sqrt{2\pi} \hat \Phi_n(k) = i^n e^{\frac{1}{2}k^2} \Big( \frac{d}{dk} \Big)^n \int_{\mathbb{R} } e^{-x^2 + \frac{1}{2}(x - ik)^2} dx = i^n \sqrt{2\pi} e^{\frac{1}{2}k^2} \Big( \frac{d}{dk} \Big)^n e^{-k^2} = (-i)^n \sqrt{2\pi} \Phi_n(k) $$

(for the second last equality we've used equation (3.14)). \( \square \)

Part b)

Let \( \chi_{[a,b]} \) be the characteristic function, \( a, b \in \mathbb{R}, a < b \). Show explicitly, i.e. without using Plancherel, that

$$ \Vert \hat \chi_{[a,b]} \Vert_2^2 = b - a. $$

Hint: You may use without proof that

$$ \int_{0}^{\infty} \frac{\sin x}{x} dx = \frac{\pi}{2} $$

Solution

Nice to know: Plancherel's theorem states that for any function \( f \in L^2: \Vert \hat f \Vert_{2} = \Vert f \Vert_{2} \) where \( \Vert f \Vert_{2} = \left( \int_{\mathbb{R}} \vert f(x) \vert^2 dx \right)^{\frac{1}{2}} \)

But we show the identity like badasses by direct calculation.

We have: $$ \hat \chi_{[a,b]} = \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} \chi_{[a,b]}(x) e^{-ikx} dx = \frac{1}{\sqrt{2\pi}} \int_{a}^b e^{-ikx} dx$$

Now we use the substitution \( y \equiv \frac{2(x-a)}{b-a} -1 \) to scale the integral to the interval \( \left[ -1, 1 \right] \). Thus:

$$ \begin{align} \hat \chi_{[a,b]} &= \frac{(b-a)}{2\sqrt{2\pi}} e^{-ik \left( \frac{1}{2}b + \frac{3}{2}a \right)}\int_{-1}^{1} e^{-ik \frac{b-a}{2} y} dy \\ &= \frac{(b-a)}{2\sqrt{2\pi}} e^{-ik \left( \frac{1}{2}b + \frac{3}{2}a \right)} \left( \frac{-2}{ik(b-a)} e^{-\frac{ik(b-a)}{2}y} \bigg \vert_{-1}^{1} \right) \\ &= \frac{1}{\sqrt{2\pi}} e^{-ik \left( \frac{1}{2}b + \frac{3}{2}a \right)} \frac{1}{ik} \left( - e^{-ik\frac{(b-a)}{2}} + e^{ik \frac{b-a}{2}} \right) \\ &= \frac{\sqrt{2}}{k\sqrt{\pi}} e^{-ik \left( \frac{1}{2}b + \frac{3}{2}a \right)} \sin \left( \frac{b-a}{2} k \right) \end{align} $$

And so:

$$ \Vert \hat \chi_{[a,b]}(k) \Vert_2^2 = \int_{\mathbb{R} } |\hat \chi_{[a,b]}(k)|^2 dk = \frac{2}{\pi} \int_{\mathbb{R} } \frac{\sin^2(\frac{b-a}{2}k)}{k^2} dk $$

With the substitution \( \eta \equiv \frac{b-a}{2}k \) we get:

$$ \Vert \hat \chi_{[a,b]}(k) \Vert_2^2 = \frac{(b-a)}{\pi} \int_{\mathbb{R} } \frac{\sin^2(\eta)}{\eta^2} d\eta $$

It remains to evaluate the integral \( I(\alpha) = \int_{\mathbb{R} } \frac{\sin^2\left(\alpha \eta\right)}{\eta^2} d\eta \) at \( \alpha = 1 \).

Amrein's Lemma: \( \frac{dI}{d\alpha} = \int_{\mathbb{R} } \frac{d}{d\alpha} \frac{\sin^2\left(\alpha \eta\right)}{\eta^2} d\eta \).

Proof of Lemma:

$$ \frac{dI}{d\alpha} = \lim_{h \rightarrow 0 } \frac{1}{h} \int_{\mathbb{R} } \frac{\sin^2 \left( (\alpha + h) \eta \right) - \sin^2 \left( \alpha \eta \right) }{\eta^2} d\eta $$

Since \( \sin(x) \leq 1 \Rightarrow \sin^2(x) \leq \vert \sin(x) \vert \) and \(\left| \sin(x) \right| \leq |x| \) (see section a), Lemma 2) we have

$$ \left| \frac{1}{h} \frac{\left| \sin \left( ( \alpha + h) \eta \right) \right| - \left| \sin \left( \alpha \eta \right) \right|}{\eta^2} \right| \leq \left| \frac{1}{h} \frac{\left| (\alpha + h)\eta\right| - \left| \alpha\eta \right| }{\eta^2} \right| \leq \left| \frac{1}{\eta} \right| $$

and with the Lebesgue dominated convergence theorem the claim holds. \( \square \)

$$ \frac{dI}{d\alpha} = 2 \int_{0}^{\infty} \frac{2 \eta \sin\left( \alpha \eta \right) \cos\left( \alpha \eta \right) }{\eta^2} $$ and with the identity $$ \sin(x) \cos(y) = \frac{1}{2}\Big(\sin (x-y) + \sin (x+y)\Big) $$ it follows that:

$$ \begin{align} \frac{dI}{d\alpha} &= 2 \int_{0}^{\infty} \frac{\sin\left( 2 \alpha \eta \right)}{\eta} d\eta \\ &= 2 \int_{0}^{\infty} \frac{\sin\left( \xi \right)}{\xi} d\xi = \pi \end{align} $$

Solving this equation is a thing we learned a long time ago:

\( I(\alpha) = \pi \alpha + c \) and since \( I(0) = 0: I(\alpha) = \pi \alpha \).

As \( I(1) = \pi \) the result is:

$$ \Vert \hat \chi_{[a,b]}(k) \Vert_2^2 = b-a $$