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Let $f\in L^1[a,b]$ satisfying $$\int^b_a t^kf(t) dt\,=0$$ for all positive integer $k$. Show that $f=0$ a.e.

I did a similar problem where $\int^b_a t^kf(t) dt\,=0$ was true for all $k\in \Bbb{N} \cup \{0\}$. This is relatively easy and I did it using Weierstrass approximation theorem, but how to do when $k\in \Bbb{N}$ and k cannot take value $0$.

saz
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Mathronaut
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2 Answers2

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The Weierstrass approximation theorem says that the polynomial functions are dense in $C[a, b]$. Since $f \in L^1[a, b]$, it is not necessarily continuous. We should note that $C[a, b]$ is dense in $L^1[a, b]$ before applying the theorem.

To answer your question, there are two approaches. The first is easier. The second is more general.

  • Consider the function $g(t) = tf(t)$. It is clear that $\int_a^b t^k g(t)\,dt = 0$ for all $k \in \{0, 1, \ldots\}$. By the result you already have, it follows that $g = 0$ [a.e.]. Thus, $f = 0$ [a.e.].
  • Use the Müntz–Szász theorem, which generalizes the Weierstrass approximation theorem.
Ayman Hourieh
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Alternative approach using Fourier transforms:

Fix $\xi \in \mathbb{R}$. By the definition of the exponential function, we have

$$\sum_{n=0}^k \frac{(\imath t \xi)^n}{n!} f(t) \to e^{\imath t \xi} f(t)$$

as $k \to \infty$ for any $t \in [a,b]$. Moreover,

$$\left|\sum_{n=0}^k \frac{(\imath t \xi)^n}{n!} f(t) \right| \leq |f(t)| e^{|\xi| \max\{|a|,|b|\}} \in L^1([a,b])$$

for all $k \geq 0$. Therefore, we conclude from the dominated convergence theorem that

$$\hat{f}(\xi := \int_a^b e^{\imath \, t \xi} f(t) \, dt = \lim_{k \to \infty} \sum_{n=0}^k \frac{(\imath \xi)^n}{n!} \int_a^b t^n \cdot f(t) \, dt =\int_a^b f(t) \, dt.$$

Since $\xi \in \mathbb{R}$ is arbitrary, this shows that the Fourier transform $\hat{f}$ of $f \cdot 1_{[a,b]}$ equals the constant $c:= \int_a^b f(t) \, dt$. On the other hand, the Riemann-Lebesgue lemma states that $\hat{f} \in C_{\infty}$; in particular

$$c=\lim_{|\xi| \to \infty} |\hat{f}(\xi)| = 0.$$

Now the uniqueness of the Fourier transform yields $f=0$ (Lebesgue-)almost everywhere on $[a,b]$.

saz
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  • Nice answer but I'm not familiar with Fourier results and I think I won't be able to use this argument in Measure Theory exam. – Mathronaut May 09 '15 at 08:23
  • @NeerajBhauryal That's a pity, but perhaps there are other readers who are interested in this question and familiar with Fourier transforms. – saz May 09 '15 at 08:27
  • Yes it might be helpful for someone,thank you for the answer. Do you know any other method? – Mathronaut May 09 '15 at 08:30
  • @NeerajBhauryal See my edited answer. – saz May 09 '15 at 08:52
  • how did you prove $\int_a^b p_k(t) f(t) , dt = c \cdot 1$? – Mathronaut May 09 '15 at 09:53
  • @NeerajBhauryal $p_k(t) := \frac{(b-t)^k}{(b-a)^k}$. To find $a_0$ just substitute in $t=0$, which doesn't give you $1$ but does give you a constant $\frac{b}{b-a}$; it doesn't really matter, since your sequence of integrands are each bounded by an integrable function (i,e, itself with integral $c. \frac{b}{b-a}$), so we can still use DCT. – Ilham May 09 '15 at 10:14