Let \(f:[0,1]\to\mathbb R\) be continuous and define \[ h(z)=\int_0^1 f(t)\cos(zt)\,dt. \]
- Prove that \(h\) is entire.
- Prove that \(h\equiv0\) only if \(f\equiv0\).
Solution
Let \[ M\coloneqq\max_{0\leq t\leq1}\abs{f(t)}. \]
1. For every \(z\in\CC\),\[ h(z) = \sum_{m=0}^{\infty} \frac{(-1)^m z^{2m}}{(2m)!} \int_0^1 f(t)t^{2m}\,dt. \]
The cosine power series gives \[ \cos(zt) = \sum_{m=0}^{\infty} \frac{(-1)^m z^{2m}t^{2m}}{(2m)!}. \] Fix \(R>0\). For \(\abs z\leq R\) and \(0\leq t\leq1\), the absolute value of the \(m\)th summand after multiplication by \(f(t)\) is at most \[ M\frac{R^{2m}}{(2m)!}. \] The series of these bounds converges. Hence the series for \(f(t)\cos(zt)\) converges uniformly for \[ (z,t)\in\{z:\abs z\leq R\}\times[0,1]. \] Termwise integration is therefore valid and gives the displayed expansion.
2. The function \(h\) is entire.
Set \[ \mu_{2m}\coloneqq\int_0^1 f(t)t^{2m}\,dt. \] Then \[ \abs{\mu_{2m}}\leq M. \] Thus the series from step 1 is dominated in absolute value by \[ M\sum_{m=0}^{\infty}\frac{\abs{z}^{2m}}{(2m)!}, \] which converges for every \(z\in\CC\). It is therefore a power series with infinite radius of convergence, so it defines an entire function. Step 1 identifies that function with \(h\).
3. If \(h\equiv0\), then\[ \int_0^1 f(t)t^{2m}\,dt=0 \]for every integer \(m\geq0\).
If \(h\) is identically zero, every coefficient of its power series at \(0\) vanishes. By step 1, the coefficient of \(z^{2m}\) is \[ \frac{(-1)^m}{(2m)!} \int_0^1f(t)t^{2m}\,dt. \] Hence every displayed moment is zero.
4. If \(h\equiv0\), then\[ \int_0^1 f(t)p(t^2)\,dt=0 \]for every real polynomial \(p\).
Every polynomial \(p(t^2)\) is a finite linear combination of \(1,t^2,t^4,\ldots\). Apply linearity to the vanishing moments in step 3.
5. There is a sequence of real polynomials \((p_j)\) such that\[ p_j(t^2)\longrightarrow f(t) \]uniformly for \(t\in[0,1]\).
Define \[ F(s)\coloneqq f(\sqrt{s}), \qquad 0\leq s\leq1. \] The function \(F\) is continuous. By the Weierstrass approximation theorem, there are real polynomials \(p_j\) converging uniformly to \(F\) on \([0,1]\). Substituting \(s=t^2\) gives the claimed uniform convergence.
6. If \(h\equiv0\), then \(f\equiv0\).
By step 4, \[ \int_0^1 f(t)p_j(t^2)\,dt=0 \] for every \(j\). Step 5 and boundedness of \(f\) imply \[ \int_0^1 f(t)p_j(t^2)\,dt \longrightarrow \int_0^1 f(t)^2\,dt. \] Therefore \[ \int_0^1 f(t)^2\,dt=0. \] The continuous function \(f^2\) is nonnegative, so a zero integral forces \(f(t)^2=0\) for every \(t\in[0,1]\). Hence \(f\equiv0\).
7. Q.E.D.
Step 2 proves part 1, and step 6 proves part 2.