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\vspace*{.3in}
\noindent{\bf Rudin, page 165/20:}

\proof
Let $P(x) = c_n x^n + \dots + c_1 x + c_0$ be a polynomial.  Then
$$
\int_0^1 f(x)P(x)\,dx =
c_n \int_0^1 f(x)x^n\,dx + \dots c_1 \int_0^1 f(x)x\,dx 
+ c_0\int_0^1 f(x)x^0\,dx = 0
$$
by hypothesis.  By Weierstrass' Approximation Theorem, there is a sequence
$\{P_n\}$ of polynomials converging uniformly to $f$ on $[0,1]$.  Moreover,
$f$ is continuous and therefore bounded on $[0,1]$, as is each of the 
polynomials $P_n$.  Therefore (see problem 2 from this section),  
$f\cdot P_n$ converges uniformly to $f\cdot f = f^2$ on $[0,1]$.  By Theorem
7.16, I conclude that 
$$
0 = \lim_{n\to\infty} \int_0^1 f(x)P_n(x) \,dx 
= \int_0^1 \lim_{n\to\infty} f(x)P_n(x)\,dx
= \int_0^1 [f(x)]^2\,dx.
$$
Now $f^2$ is a non-negative function that vanishes exactly where 
$f$ does.  So if $f(x)>0$ for some $x\in [0,1]$, it follows from continuity
of $f$ that for some $\delta>0$, $|t-x|<\delta$ implies that
$f(t) > f(x)/2$.  Therefore,
$$
\int_0^1 [f(x)]^2\,dx \geq \delta [f(x)]^2/4 > 0.
$$
This is impossible, so $f\equiv 0$ on $[0,1]$.
\qed
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