Assume that $ f ∈ L([a, b])$ and $\int x^nf(x)dx=0$ for $n=0,1,2...$.

Assume that $ f ∈ L([a, b])$ and $\int x^nf(x)dx=0$ for $n=0,1,2...$. Prove $f=0$ a.e. Since there exist polynomials going to f almost everywhere all I would need to do is bring the limit in to prove that $ ||f||_2 = 0$ and then I'm done. But no success with this. Thanks for any help.

Solution 1:

Byron's answer is perfectly OK. Here is another one.

Since the polynomials are uniformly dense in $\mathcal C([a,b])$ (the space of continuous functions on $[a,b]$), it follows from the assumption that $\int_{[a,b]} g(x)f(x)\, dx=0$ for all $g\in\mathcal C([a,b])$. (This means that the signed measure $f(x)dx$ is the $0$ measure). From this, "it is well known" that one can deduce that $f=0$ a.e. The details are as follows.

Assume that $f$ is not equal to $0$ a.e. Then, for example, the set $\{ f>0\}$ has positive measure. So one can find $\varepsilon >0$ such that the set $\{ f>\varepsilon\}$ has positive measure. By the regularity of the Lebesgue measure, one can find a compact set $K$ and an open set $V$ such that $K\subset \{ f>\varepsilon\}\subset V$ and the measure of $V\setminus K$ is as small as we wish. Now, choose a continuous function $g$ such that $0\leq g\leq 1$ which is equal to $1$ on $K$ and to $0$ outside $V$. Then $$\left\vert\int_{[a,b]} gf\right\vert=\left\vert\int_V gf\right\vert\geq \left\vert\int_Kgf\right\vert-\int_{V\setminus K}\vert gf\vert\geq \varepsilon\, m(K)-\int_{V\setminus K} \vert f\vert\, , $$ where $m(K)$ is the measure of $K$. Since the measure of $V\setminus K$ is as small as we wish, we may assume that $\int_{V\setminus K} \vert f\vert<\varepsilon\, m(K)$, and it follows that $\int_{[a,b]} gf\neq 0$; which is a contradiction.

Solution 2:

A modification of my answer here.

You could use the functional form of the Monotone Class Theorem. Let $\cal H$ be the collection of all bounded, Borel measurable functions $g$ so that $\int g(x) f(x)\, dx=0$. Then $\cal H$ is a monotone vector space (exercise for the reader!).

Let $\cal K$ be the set of functions $\{x^k: k\in\mathbb{N}\}$. Then $\cal K$ is a multiplicative class contained in $\cal H$, so the Monotone Class Theorem says that $$b(\sigma({\cal K}))\subseteq {\cal H},$$ where $b(\sigma({\cal K}))$ is the space of all bounded functions measurable with respect to the $\sigma$-algebra generated by $\cal K$. Since $\cal K$ generates the Borel $\sigma$-algebra we deduce that $\mbox{sgn}(f)\in{\cal H}$ and hence that $\int\mbox{sgn}(f(x)) f(x)\,dx= \int |f(x)|\, dx=0$.

Solution 3:

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

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

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

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

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

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

Since $\xi \in \mathbb{R}$ is arbitrary, this shows that the Fourier transform of $f \cdot 1_{[a,b]}$ equals $0$. Now the uniqueness of the Fourier transform yields $f=0$ (Lebesgue-)almost everywhere on $[a,b]$.

Remark: This question shows that it suffices to assume $\int_a^b x^n f(x) \, dx =0$ for all $n \geq 1$.

Solution 4:

Define $$ F(x)=\int_a^xf(t)\,\mathrm{d}t\tag{1} $$ Since $f\in L^1([a,b])$, $F$ is continuous.

By definition, $F(a)=0$. Furthermore, by hypothesis, $\displaystyle F(b)=\int_a^bf(t)\,\mathrm{d}t=0$.

For any polynomial, $p$, let $\displaystyle P(x)=\int_a^xp(t)\,\mathrm{d}t$, which is also a polynomial. Then integration by parts yields $$ \begin{align} \int_a^bF(x)p(x)\,\mathrm{d}x &=F(b)P(b)-F(a)P(a)-\int_a^bf(x)P(x)\,\mathrm{d}x\\ &=0\tag{2} \end{align} $$ The Stone-Weierstrass Theorem says that polynomials are dense in $C([a,b])$. Thus, for any $\epsilon\gt0$, there is a polynomial, $p$, so that $$ \max_{x\in[a,b]}|F(x)-p(x)|\le\epsilon\tag{3} $$ Thus, $$ \int_a^b(F(x)-p(x))^2\,\mathrm{d}x\le(b-a)\epsilon^2\tag{4} $$ Furthermore, using $(2)$, we get $$ \begin{align} \int_a^b(F(x)-p(x))^2\,\mathrm{d}x &=\int_a^bF(x)^2\,\mathrm{d}x+\int_a^bp(x)^2\,\mathrm{d}x-2\int_a^bF(x)p(x)\,\mathrm{d}x\\ &=\int_a^bF(x)^2\,\mathrm{d}x+\int_a^bp(x)^2\,\mathrm{d}x\tag{5} \end{align} $$ $(4)$ and $(5)$ imply that $$ \int_a^bF(x)^2\,\mathrm{d}x\le(b-a)\epsilon^2\tag{6} $$ Since $\epsilon\gt0$ was arbitrary, we must have $$ \int_a^bF(x)^2\,\mathrm{d}x=0\tag{7} $$ Since $F$ is continuous, $(7)$ implies that $F(x)=0$ for $x\in[a,b]$. Thus $(1)$ implies that $$ f(x)=0\quad\text{a.e.}\tag{8} $$

Solution 5:

By hypothesis is easy to conclude that $\int pf \ dx = 0$ for all polynomials $p$ (*). Consider $\varphi $ a smooth function with compact support. By the Weierstrass approximation theorem exists a sequence $p_n$ of polynomial that converges uniformly to $\varphi$. Then given $\epsilon >0$ exists $n_o \in N$ where $|p_n (x) - \varphi (x)| < \epsilon$ for all $x \in [a,b]$ and for all $n \geq n_o$ .

We have for $n \geq n_o$ fixed:

$$ |\int \varphi f| = |\int \varphi f - p_nf + p_nf| \leq |\int (\varphi - p_n)f| + |\int p_nf| = |\int (\varphi - p_n)f | (by \ (*) )$$

$$ \leq \int |(\varphi - p_n)f| \leq \epsilon \int |f|$$

Then we conclude that $\int \varphi f =0 $ . By the du Bois-Reymond theorem (see theorem 17 of http://people.oregonstate.edu/~peterseb/mth627/docs/627w2004-convolution.pdf) we can conclude that $f=0$.