Consider a complex-valued function of the form $$ \newcommand{\bfx}{\mathbf{x}} \newcommand{\bfk}{\mathbf{k}} f(t,\bfx)=\int d^Nk\ g(\bfk)\exp\big(-i\omega(\bfk)t-i\bfk\cdot\bfx\big) \tag{1} $$ where boldface denotes a list of $N$ real variables, the dot-product is defined as usual, and $$ \omega(\bfk)\equiv \sqrt{\strut{}1+\bfk\cdot\bfk}. \tag{2} $$ (I'm calling this a "wave," but notice the constant term under the square root.) Suppose that $f(0,\bfx)$ is nonzero at least for some $\bfx$. Can we choose $g(\bfk)$ so that $f(t,\bfx)$ and $df(t,\bfx)/dt$ both have compact support in $\bfx$ at $t=0$?

The answer must be no, because otherwise I could use the Paley-Wiener theorem to construct a contradiction to the Reeh-Schlieder theorem. But that's a very indirect argument that uses relativistic quantum field theory, which surely isn't necessary for the simple question I'm asking here! How can we prove more directly that no such $g(\bfk)$ exists?

  • $\begingroup$ I am trying to understand the same issue but for more general functions, since recently I found the paper Finite Time Differential Equations I am trying to figure out if their findings can be extended to more than one dimensions (unsuccessfully so far). Maybe it could help you. $\endgroup$
    – Joako
    Feb 21, 2022 at 2:52
  • $\begingroup$ I am not fully sure about this, but I think it is possible to adjust the function shown in this answer to a question I made to be a counter-example to your question. $\endgroup$
    – Joako
    Mar 27 at 2:11
  • $\begingroup$ @Joako Thank you for the comment! I don't know how to adjust that example to construct a counter-example, though. If we omit the "$1$" term under the square root in my equation (2), then any function of the form $f(kx-ct)$ with compact support at $t=0$ would be a counter-example with $N=1$. It's that pesky "$1$" term under the square root that makes things difficult. $\endgroup$ Mar 28 at 23:29
  • $\begingroup$ @Joako By the way, the function $E(x,y,z,t)=f(k_xx-ct, k_yy-ct, k_zz-ct)$ shown in that example doesn't satisfy the usual wave equation. Not sure what the author of that answer meant. $\endgroup$ Mar 28 at 23:29
  • 1
    $\begingroup$ @Joako Now let $f(x_1,...,x_N)$ be any smooth function, and suppose $$ E(t,x_1,...,x_N)=f(x_1-t,x_2-t,...,x_N-t). $$ Any such function $E(t,x_1,...,x_N)$ satisfies $$ \partial_t^2 E = (\partial_1+\partial_2+\cdots+\partial_N)^2 E. $$ If $E$ also satisfies the wave equation, then this immediately implies $$ \sum_{j\neq k} \partial_j\partial_k E = 0. $$ By inspection, the example described in the other answer doesn't satisfy this when $N\geq 2$, not even when $t=0$. $\endgroup$ Mar 29 at 2:29


You must log in to answer this question.