# If $f \in C[a,b]$ has $\int_{a}^{b}f(x)x^{n}dx=0$ for all $n\in \mathbb{N}$, then $f=0$.

Let $$f \in C[a,b]$$ with $$\int_{a}^{b}f(x)x^{n}dx=0$$ for all $$n\in \mathbb{N}$$. Prove $$f=0$$.

I got the intuition to prove this with induction over $$n \in \mathbb{N}$$, for $$n=0$$, I have $$\int_{a}^{b}f(x)dx=0$$. So how I got that $$f=0$$? Also, how I end up the proof? Any help will be appreciated. Thanks

• Induction isn't the right approach here, I'm afraid. – Jakobian Feb 27 at 1:33
• Any other idea? @Jakobian – Cos Feb 27 at 1:36
• A quite similar question was asked about a day ago at Use Weierstrass to show $f(x)=0$. It asks a question of the same form as yours, except it's restricted to $a = 0$, $b = 1$, and $n$ being even. Nonetheless, you may find the answers there helpful to you. – John Omielan Feb 27 at 1:36

Based on the context set in the text of the question, I assume we can take

$$0 \in \Bbb N; \tag 1$$

I also take it that

$$a \le b. \tag 2$$

We have the following

Lemma: Suppose

$$g_n \in C[a, b] \tag 3$$

is a sequence such that

$$g_n \to f \; \text{as} \; n \to \infty \tag 4$$

then

$$\displaystyle \lim_{n \to \infty} \int_a^b fg_n \; dx = \int_a^b f^2 \; dx. \tag 5$$

Proof of Lemma:

Observe that

$$\displaystyle \int_a^b f^2 \; dx - \int_a^b f g_n \; dx = \int_a^b (f^2 - fg_n) \; dx = \int_a^b f(f - g_n) \; dx; \tag 7$$

thus

$$\left \vert \displaystyle \int_a^b f^2 \; dx - \int_a^b f g_n \; dx \right \vert = \left \vert \displaystyle \int_a^b f(f - g_n) \; dx \right \vert$$ $$\le \displaystyle \int_a^b \vert f (f - g_n) \vert \; dx = \displaystyle \int_a^b \vert f \vert \vert f - g_n \vert \; dx \le \int_a^b \vert f \vert \Vert f - g_n \Vert \; dx = \Vert f - g_n \Vert \int_a^b \vert f \vert \; dx; \tag 8$$

passing to the limit,

$$\lim_{n \to \infty} \left \vert \displaystyle \int_a^b f^2 \; dx - \int_a^b f g_n \; dx \right \vert \le \lim_{n \to \infty} \Vert f - g_n \Vert \displaystyle \int_a^b \vert f \vert \; dx = 0; \tag 9$$

thus (5) binds. End: Proof of Lemma.

With this lemma in hand we proceed with the observation that the function

$$x:[a, b] \to \Bbb R \tag{10}$$

separates points in $$[a, b]$$; that is, for any

$$c_1, c_2 \in [a, b], \; c_1 \ne c_2, \tag{11}$$

we have

$$x(c_1) = c_1 \ne c_2 = x(c_2); \tag{12}$$

it then follows from the Stone-Weierstrass theoremthat the sub-algebra of $$C[a, b]$$ generated by $$x$$ and the constant functions is dense; since this sub-algebra is the set of real polynomials in $$x$$, it follows that there exists a sequence of polynomials $$p_n(x)$$ such that

$$p_n(x) \to f(x) \; \text{in} \; C[a, b]; \tag{13}$$

that is,

$$\displaystyle \lim_{n \to \infty} \Vert p_n(x) - f(x) \Vert = 0, \tag{14}$$

and from the hypothesis that

$$\displaystyle \int_a^b x^m f(x) \; dx = 0, \; \forall m \in \Bbb N, \tag{15}$$

the linearity of the integral allows us to conclude that

$$\displaystyle \int_a^b p_n(x) f(x) \; dx = 0, \forall n \in \Bbb N; \tag{16}$$

now applying the lemma we find

$$\left \vert \displaystyle \int_a^b f^2(x) \; dx \right \vert = \lim_{n \to \infty} \left \vert \displaystyle \int_a^b f^2 \; dx - \underbrace{\int_a^b p_n f \; dx}_{0} \right \vert \le \lim_{n \to \infty} \Vert f - p_n \Vert \displaystyle \int_a^b \vert f \vert \; dx = 0, \tag{17}$$

which forces

$$\displaystyle \int_a^b f^2(x) \; dx = 0; \tag{18}$$

now since $$f^2(x)$$ is a non-negative continuous function it follows that

$$f(x) = 0, \forall x \in [a, b]. \tag{19}$$

$$OE\Delta$$.

Use the fact that polynomials are dense in $$C[a,b]$$ with the supremum norm (see the Stone–Weierstrass theorem).

Due to the linearity of integration, your assumption implies that

$$\int_a^b f(x)P(x) \mathrm{d}x = 0 \hspace{10px}(\star)$$

for any polynomial $$P(x)$$.

By the Stone-Weierstrass theorem, you can find a sequence of polynomials $$\{P_n\}_{n=1}^{\infty}$$ such that $$\lim_{n\to\infty}P_n(x) = f(x)$$ uniformly. Since the convergence is uniform, we can exchange limit and integration operators. Therefore, we have

$$\lim_{n\to\infty}\underbrace{\int_a^b f(x)P_n(x)\mathrm{d}x}_{\text{this is zero by }(\star)} = \int_a^bf(x)\lim_{n\to\infty}P_n(x)\mathrm{d}x= \int_a^b f(x)^2 \mathrm{d}x = 0$$

Since $$f^2(x) \geq 0$$, we conclude that $$f=0$$ on $$[a,b]$$.