Let \(f:[0,1]\to\mathbb C\) be continuous. Show that \[ g(z)=\int_0^1 f(t)e^{tz^2}\,dt \] defines an entire function \(g\) on \(\mathbb C\).
Solution
Since \(f\) is continuous on \([0,1]\), set \[ M\coloneqq\max_{0\leq t\leq1}\abs{f(t)}. \]
1. For every \(R>0\), the series\[ e^{tz^2} = \sum_{n=0}^\infty\frac{t^n z^{2n}}{n!} \]converges uniformly for\[ 0\leq t\leq1, \qquad \abs{z}\leq R. \]
For such \(t\) and \(z\), \[ \abs{\frac{t^n z^{2n}}{n!}} \leq \frac{R^{2n}}{n!}. \] Since \[ \sum_{n=0}^\infty\frac{R^{2n}}{n!} = e^{R^2}, \] the Weierstrass \(M\)-test gives uniform convergence.
2. For every \(z\in\CC\),\[ g(z) = \sum_{n=0}^\infty c_n z^{2n}, \qquad c_n\coloneqq\frac1{n!}\int_0^1 f(t)t^n\,dt. \]
Fix \(R>\abs{z}\). By step 1, the exponential series converges uniformly in \(t\in[0,1]\) for this \(z\). Multiplication by the bounded function \(f\) preserves uniform convergence, so termwise integration gives \[ \begin{aligned} g(z) &= \int_0^1 f(t) \sum_{n=0}^\infty\frac{t^n z^{2n}}{n!} \,dt\\ &= \sum_{n=0}^\infty \left( \frac1{n!}\int_0^1f(t)t^n\,dt \right)z^{2n}. \end{aligned} \]
3. The power series in step 2 has infinite radius of convergence.
For every \(n\geq0\), \[ \abs{c_n} \leq \frac{M}{n!}\int_0^1t^n\,dt = \frac{M}{(n+1)n!} \leq \frac{M}{n!}. \] Therefore, for every \(z\in\CC\), \[ \sum_{n=0}^\infty \abs{c_n z^{2n}} \leq M\sum_{n=0}^\infty\frac{\abs{z}^{2n}}{n!} = Me^{\abs{z}^2}<\infty. \] Hence the series converges everywhere in \(\CC\) and defines an entire function.
4. The function \(g\) is entire on \(\CC\).
3. Thus \(g\) is entire.
5. Q.E.D.
Step 4 is the required conclusion.