$\int_{0}^{1}\phi(t)e^{-at}dt =0$ for every $a>0$ implies $\int_{0}^{1}\phi(t)t^ndt =0.$

Question

Suppose $\phi: [0,1]\to \mathbb{R}$ is continuous, and that for every $a>0$ we have $\displaystyle \int_{0}^{1}\phi(t)e^{-at}dt =0$ then show that for every non-negative integer $n$ we have $\displaystyle \int_{0}^{1}\phi(t)t^ndt =0.$

My attempt: Let $f_n(t)= \phi(t)e^{-\frac{t}{n}}$ so that $f(t) = \displaystyle \lim_{n \to \infty}f_n(t)= \phi(t).$ Also, for every $n \in \mathbb{N},$ and $t\in [0,1]$, it's clear that $|f_n(t)|\leq |\phi(t)|,$ which is an integrable function on $[0,1]$ as it is continuous. Then by the dominated convergence theorem, we have $0=\displaystyle \lim_{n \to \infty}\int_{0}^{1}f_n(t)dt= \displaystyle \int_{0}^{1}\phi(t)dt.$

Let $F(x)= \displaystyle \int_{0}^{x}\phi(t)dt$ then since $\phi$ is continuous, by the fundamental theorem of calculus, $F$ is differentiable and $F'(x)= \phi(x)$ for all $x \in [0,1].$ Also from above, $F(1)=0.$

For non-zero $a$, integrating $\displaystyle \int_{0}^{1}\phi(t)e^{-at}dt$ by parts we get

$\begin{align}0&= \displaystyle \int_{0}^{1}\phi(t)e^{-at}dt\\&=e^{-at} F(t)\bigg|_{0}^{1} +a \displaystyle \int_{0}^{1}F(t)e^{-at}dt\end{align}$

$\Rightarrow \displaystyle \int_{0}^{1}F(t)e^{-at}dt =0$

Since $a>0$ was arbitrary it follows by the same dominated convergence argument used above that $\displaystyle \int_{0}^{1}F(t)dt =0.$ Using integration by parts once more we get $\displaystyle \int_{0}^{1}F(t)dt = -\displaystyle \int_{0}^{1}t\phi(t)dt$ so $\displaystyle \int_{0}^{1}t\phi(t)dt=0.$

Now setting $G(x)= \displaystyle \int_{0}^{x}F(t)dt$ and repeating the above argument on $G$ in place of $F$ we get $\displaystyle \int_{0}^{1}tF(t)dt=0.$ Since $\displaystyle \int_{0}^{1}tF(t)dt = -2\displaystyle \int_{0}^{1}t^2\phi(t)dt$ it follows that $\displaystyle \int_{0}^{1}t^2\phi(t)dt=0.$

Continuing this process for successive anti-derivatives of $\phi$ it follows that $\displaystyle \int_{0}^{1}t^n\phi(t)dt=0$ for all $n \in \mathbb{Z}_{\geq 0}.$

Question: Is this proof correct? If no, what is the flaw? If yes, is there a method that does not proceed case-by-case?

Thanks in advance.

Answer 1

Your proof seems fine. So let me introduce two more proofs, where one is elementary but problem-specific and the other is more advanced but robust.

---

1^st Proof. By the Taylor's theorem, for $n \geq 0$

$$ \forall a > 0, \ \forall t \geq 0 \ : \quad \left| e^{-at} - \sum_{k=0}^{n} \frac{(-1)^k}{k!} (at)^k \right| \leq \frac{(at)^{n+1}}{(n+1)!}. $$

Now define $P_n(a) := \sum_{k=0}^{n} \frac{(-1)^k}{k!} \left( \int_{0}^{1} t^k \phi(t) \, dt \right) a^k$. Then $P_n$ is a polynomial of degree at most $n$. Moreover, by the identity $\int_{0}^{1}\phi(t)e^{-at} \, dt = 0$, we find that for $a > 0$,

$$ \left| P_n(a) \right| = \left| \int_{0}^{1} \phi(t) \left( e^{-at} - \sum_{k=0}^{n} \frac{(-1)^k}{k!} (at)^k \right) \, dt \right| \leq \frac{a^{n+1}}{(n+1)!} \int_{0}^{1} t^{n+1} \lvert \phi(t) \rvert \, dt. $$

This is enough to conclude that $P_n \equiv 0$ and hence $\int_{0}^{1} t^k \phi(t) \, dt = 0$ for $k = 0, \cdots, n$. (See Lemma below, for instance.) But since $n$ was arbitrary, this is true for all $k \geq 0$. ////

• Lemma. Let $P(x) = \sum_{k=0}^{n} c_k x^k$ be a polynomial of degree at most $n$. If there exists $C \geq 0$ such that $\lvert P(x) \rvert \leq Cx^{n+1}$ for all $x > 0$, then $c_k = 0$ for each $k = 0, \cdots, n$.

  Proof of Lemma. We obtain $a_0 = 0$ from the computation

  $$ 0 \leq \left| a_0 \right| = \lim_{x \to 0^+} \left| P(x) \right| \leq \lim_{x\to0^+} Cx^{n+1} = 0. $$

  Likewise, if we have discovered that $a_0 = \cdots = a_{k-1} = 0$ for $k \in \{1, \cdots, n\}$, then

  $$ 0 \leq \left| a_k \right| = \lim_{x \to 0^+} \frac{\left| P(x) \right|}{x^k} \leq \lim_{x\to0^+} Cx^{n+1-k} = 0 $$

  and hence $a_k = 0$. Therefore the claim follows by induction on $k$. ////

---

2^nd Proof. We prove the following claim, which is enough for initiating induction in OP's case.

Theorem. Let $\varphi : [0,\infty) \to \mathbb{R}$ be a measurable function such that $t \mapsto \varphi(t)e^{-at}$ is integrable on $[0,\infty)$ for each $a > 0$. Define

$$L(a) := \int_{0}^{\infty} \varphi(t)e^{-at} \, dt. $$

Then

1. $L$ is differentiable on $(0, \infty)$ with the derivative $L'(a) = - \int_{0}^{\infty} t \varphi(t)e^{-at} \, dt$.

2. If $\varphi$ is integrable, then $L(a) \to L(0)$ as $a \to 0^+$.

Proof. Part 2 is a straightforward application of the dominated convergence theorem. Assume that $\varphi$ is integrable. Then $t \mapsto \varphi(t)e^{-a t}$ is dominated by $\lvert \varphi(t) \rvert$. So $L(a_n) \to L(0)$ as $n\to\infty$ along any sequence $(a_n)$ of positive reals such that $a_n \to 0$ as $n\to \infty$. Therefore $L(a) \to L(0)$ as $a \to 0^+$.

Next we prove Part 2. Write $G(a) = - \int_{0}^{\infty} t \varphi(t)e^{-at} \, dt$. This is well-defined since $t \mapsto t \varphi(t)e^{-at}$ is integrable on $[0, \infty)$ for each $a > 0$. (To argue this, simply use the bound $t \leq (2/a)e^{at/2}$.) Now fix $a > 0$. Then for any $h$ with $0 < \lvert h \rvert < a/2$,

\begin{align*} \left| \frac{L(a+h) - L(a)}{h} - G(a) \right| &\leq \int_{0}^{\infty} \underbrace{ \lvert \varphi(t) \rvert e^{-at} \left| \frac{e^{-ht} - 1}{h} + t \right| }_{(*)} \, dt \end{align*}

We make some observations on the integrand $\text{(*)}$.

• By the mean-value theorem, $ \left| \frac{e^{-ht} - 1}{h} \right| \leq t e^{at/2} $ and hence $\text{(*)}$ is dominated by the integrable function $ t \lvert \varphi(t) \rvert e^{-at} (e^{at/2}+1) $.

• As $h\to 0$, $\text{(*)}$ converges pointwise to $0$.

Thus along any sequence $(h_n)$ with $0 < \lvert h_n \rvert < a/2$ and $h_n \to 0$, we have $\lim_{n\to\infty} \int_{0}^{\infty} \text{(*)} \, dt = 0$ by the dominated convergence theorem. This implies that

$$ \lim_{n\to\infty} \frac{L(a+h_n) - L(a)}{h_n} = G(a). $$

Since this is true for all such $(h_n)$, it follows that $\frac{L(a+h) - L(a)}{h} \to G(a)$ as $h \to 0$. ////