Suppose that \(f\) is a continuous function on \(\RR\) which vanishes outside some finite interval and for each \(z \in \mathbb { C }\) define
\[ g ( z ) = \int _ { - \infty } ^ { \infty } f ( t ) \exp ( - i z t ) d t . \]
Show that \(g\) is entire.
Solution
Choose \(A>0\) such that \[ f(t)=0 \] whenever \[ \abs t>A. \] Then \[ g(z) = \int_{-A}^{A}f(t)e^{-izt}\,dt. \]
1. Fix \(z_0\in\CC\). For \(h\neq0\),\[ \frac{g(z_0+h)-g(z_0)}h = \int_{-A}^{A} f(t)e^{-iz_0t} \frac{e^{-iht}-1}{h} \,dt. \]
By the compact-support reduction above, \[ \begin{aligned} g(z_0+h)-g(z_0) &= \int_{-A}^{A} f(t) \bigl(e^{-i(z_0+h)t}-e^{-iz_0t}\bigr) \,dt\\ &= \int_{-A}^{A} f(t)e^{-iz_0t} \bigl(e^{-iht}-1\bigr) \,dt. \end{aligned} \] Divide by the nonzero complex number \(h\).
2. As \(h\to0\),\[ \frac{e^{-iht}-1}{h} \longrightarrow -it \]uniformly for\[ t\in[-A,A]. \]
Define the entire function \[ \Phi(w) = \begin{cases} \dfrac{e^w-1}{w},&w\neq0,\\ 1,&w=0. \end{cases} \] Then \[ \frac{e^{-iht}-1}{h} = -it\,\Phi(-iht). \] For \[ \abs t\leq A, \] one has \[ \abs{-iht}\leq A\abs h\longrightarrow0 \] uniformly in \(t\). Since \(\Phi\) is continuous at \(0\), \[ \sup_{\abs t\leq A} \abs{\Phi(-iht)-1} \longrightarrow0. \] Multiplying by \(\abs t\leq A\) gives the claimed uniform convergence.
3. The integrands in step 1 converge uniformly on \([-A,A]\) to\[ -it\,f(t)e^{-iz_0t}. \]
The continuous function \[ t\longmapsto f(t)e^{-iz_0t} \] is bounded on the compact interval \([-A,A]\). Multiplying the uniform convergence from step 2 by this bounded factor preserves uniform convergence.
4. The complex derivative of \(g\) exists at \(z_0\) and is\[ g'(z_0) = \int_{-A}^{A} (-it)f(t)e^{-iz_0t}\,dt. \]
1. Hence\[ \begin{aligned} g'(z_0) &= \lim_{h\to0} \frac{g(z_0+h)-g(z_0)}h\\ &= \int_{-A}^{A} (-it)f(t)e^{-iz_0t}\,dt. \end{aligned} \]
5. The function \(g\) is entire.
The point \(z_0\in\CC\) was arbitrary. Step 4 shows that the complex derivative exists at every point of \(\CC\). Therefore \(g\) is entire.
6. Q.E.D.
Step 5 is the required conclusion.