Variant of Baby Rudin problem 7.20

Question

I am attempting to prove the following, which is a variant of Baby Rudin problem 7.20:

Let $f:[0,1]\to\mathbb R$ be a Riemann-integrable function with $|f(x)|\le1$ for all $x\in[0,1]$. Suppose

$$\int_0^1 f(x)x^n dx=0$$

for each $n=0,1,2,\dots$. Assume $f$ is continuous at some $x_0\in[0,1]$. Show $f(x_0)=0$.

I am unsure where to start. Here are some of my ideas for the proof to get the ball rolling:

Assume to the contrary that $f(x_0)\ne0$.

Define Euclidean distance:

$$||f||_2=\left(\int_0^1 |f(x)|^2dx\right)^\frac12.$$

First, we need to show that $\int_0^1 |f(x)|^2dx=0$. Fix arbitrary $\epsilon>0$. According to Baby Rudin problem 6.12, there exists some continuous $g:[0,1]\to\mathbb R$ such that $||f-g||_2<\epsilon$. I believe it is true that $||f||_2\le||f-g||_2+||g||_2<\epsilon$. The problem is that I need to justify that $\int_0^1 g(x)x^ndx=0$ for any $n\ge0$ before applying the result from problem 7.20 to conclude that $||g||_2=0$. If this conjecture is true, then I will successfully prove $||f||_2=0$.

Finally, I can use the fact that $f$ is continuous at $x_0$ to show that $\int_0^1|f(x)|^2dx>0$, which contradicts with $||f||_2=0$.

Will it work?

Answer 1

If $f$ is continuous in at $x_0$ and suppose for the sake of contradiction that $f(x_0) \neq 0$, then there exists $\epsilon>0$ such that $|f|$ is greater than $\epsilon$ for some open interval $(x_0-r,x_0+r)$ (WLOG we can take $f$ non negative in this region by considering $-f$ if it isn't). Approximate the indicator function $\chi_{[x_0-r,x_0+r]}(x)$ by a polynomial $p(x)$ in $L^1$ norm up to $\delta$ for arbitrary $\delta>0$ (here we use the fairly strong statement that polynomials are dense in $L^1$), then we have: $$\|\chi_{[x_0-r,x_0+r]} - p\|_1< \delta \,\,\textrm{ and }\,\, \int_0^1 (fp)(x) dx = 0$$ And so: $$ \int_0^1f(x)\chi_{[x_0-r,x_0+r]}(x)dx= \int_{x_0-r}^{x_0+r}f(x)dx \geq \int_{x_0-r}^{x_0+r}\epsilon dx \geq 2r\epsilon$$ Yet: $$ \int_0^1f(x)\chi_{[x_0-r,x_0+r]}(x)dx=\int_0^1f(x)\chi_{[x_0-r,x_0+r]}(x)dx - \int_0^1 f(x) p(x) dx$$ $$\leq \int_0^1 |f(x)||\chi_{[x_0-r,x_0+r]}(x)-p(x)| dx \leq 1 \int_0^1|\chi_{[x_0-r,x_0+r]}(x)-p(x)| dx\leq \delta$$ Take $\delta \to 0$ and we have the contradiction $2\epsilon r \leq 0$.