Web Bonus Problem to Week 8 (#1)

[Victor Ivrii]

Problem 6

http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter5/S5.2.P.html#problem-5.2.P.6

[Jeremy Li 2]

PART A
In case the boldfacing isn't clear below, $\mathbf{k}$ and $\mathbf{x}$ are vectors.

a)
Let $\kappa$ be $\sqrt{2\pi}$. Then, applying the Plancherel theorem to each dimension:
\begin{equation}
||Fu(\mathbf{x})||=||\hat{u}(\mathbf{x})||=||u(\mathbf{x})||
\end{equation}

Then $F^*F=I$.

This is an infinite dimensional operator, but right-multiplying the previous equation by $F^{-1}$ we get $F^*=F^{-1}$. Therefore, $FF^*=FF^{-1}=I$

b)
\begin{equation}
(F^2 f)(\mathbf{x}) = F(F(f(\mathbf{x})))=F(\hat{f}(\mathbf{k}))
=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} \hat{f}(\mathbf{k}) e^{-i\mathbf{k}\cdot\mathbf{x}}d^n\mathbf{k}
=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} \hat{f}(\mathbf{k}) e^{i\mathbf{k}\cdot(-\mathbf{x})}d^n\mathbf{k}
=f(-\mathbf{x})
\end{equation}

c)
This is obvious from b.
\begin{equation}
(F^4f)(\mathbf{x})=(F^2F^2f)(\mathbf{x})=(F^2f)(-\mathbf{x})=f(\mathbf{x})
\end{equation}

d)
Firstly, assume $f(\mathbf{x})$ is even,

Then
\begin{equation}
\hat{f}(\mathbf{k})=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{x}) e^{-i \mathbf{k} \cdot \mathbf{x}} d^n\mathbf{x}
\end{equation}

Expanding the complex exponential:
\begin{equation}
\hat{f}(\mathbf{k})=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{x}) \cos(\mathbf{k}\cdot\mathbf{x}) d^n\mathbf{x} - \frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{x}) i\sin(\mathbf{k}\cdot\mathbf{x}) d^n\mathbf{x}
\end{equation}

Since $f(\mathbf{x})$ is even, we can repeat the calculation of the Fourier Transform with $f(-\mathbf{x})$ instead.
\begin{equation}
\hat{f}(\mathbf{k})=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{-x}) \cos(\mathbf{k}\cdot\mathbf{x}) d^n\mathbf{x} - \frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{-x}) i\sin(\mathbf{k}\cdot\mathbf{x}) d^n\mathbf{x}
\end{equation}

We can do the substitution $\mathbf{-x}=\mathbf{x}$. We get a negative factor in the differential when we do this, but the limits of integration are also reversed on each coordinate, so the negative factor cancels. Therefore:
\begin{equation}
\hat{f}(\mathbf{k})=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{x}) \cos(-\mathbf{k}\cdot\mathbf{x}) d^n\mathbf{x} - \frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{x}) i\sin(-\mathbf{k}\cdot\mathbf{x}) d^n\mathbf{x}\\
=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{x}) \cos(\mathbf{k}\cdot\mathbf{x}) d^n\mathbf{x} + \frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{x}) i\sin(\mathbf{k}\cdot\mathbf{x}) d^n\mathbf{x}
\end{equation}

Adding (5) and (7) and dividing by 2 we get:
\begin{equation}
\hat{f}(\mathbf{k})=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{x}) \cos(\mathbf{k}\cdot\mathbf{x}) d^n\mathbf{x}
\end{equation}
which is real if $f(\mathbf{x})$ is real.

Now assume that $\hat{f}(\mathbf{k})$ is real. We want to prove that $f(\mathbf{x})$ is even. The IFT is:
\begin{equation}
f(\mathbf{x})=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} \hat{f}(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x}} d^n\mathbf{k}
\end{equation}

Expanding the complex exponential
\begin{equation}
f(\mathbf{x})=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} \hat{f}(\mathbf{k}) \cos(\mathbf{k} \cdot \mathbf{x}) d^n\mathbf{k} + \frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} \hat{f}(\mathbf{k}) i\sin(\mathbf{k} \cdot \mathbf{x}) d^n\mathbf{k}
\end{equation}

If $\hat{f}(\mathbf{k})$ is real, than the integrand of the term on the right is purely imaginary, whereas the term on the left is purely real. With the given assumption that $f(\mathbf{x})$ is real, the second term must necessarily be zero. Then:

\begin{equation}
f(\mathbf{x})=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} \hat{f}(\mathbf{k}) \cos(\mathbf{k} \cdot \mathbf{x}) d^n\mathbf{k}
\end{equation}

Therefore $f(\mathbf{x})$ is even.

The case of odd $f(\mathbf{x})$ can be done in the same way.

[Jeremy Li 2]

PART B
The transformation $Q$ is orthogonal. That means $Q^T=Q^{-1}$.

Then
\begin{equation}
\det(Q)\det(Q)=\det(Q)\det(Q^T)=\det(QQ^T)=\det(QQ^{-1})=\det(I)=1
\end{equation}

This means that $|\det(Q)|=1$. By 5.2.A Eq(1):
\begin{equation}
f(\mathbf{x})=f(Q\mathbf{x}) \iff \hat{f}(\mathbf{k})=\hat{f}(Q^T\mathbf{k})
\end{equation}

It's evident that if $f$ is rotationally invariant (left side is true), than the right side holds for any $Q^T$ orthogonal, and so $\hat{f}$ is rotationally invariant. It's evident that if $\hat{f}$ is rotationally invariant (right side is true), than the left side holds for any $Q$ orthogonal, and so $f$ is rotationally invariant.