# Counterexample to the set of all algebraic polynomials being dense in $[0,1]$

Today in class the professor said that if we consider $X = [0, 1]$ to be equipped with the Lebesgue measure, then the set of all algebraic polynomials is not dense in $L^\infty([0, 1])$. But I couldn't come up with a counterexample to convince myself.

Could someone come up with one?

• What is an algebraic polynomial? – Rudy the Reindeer Nov 5 '14 at 0:37
• It is a polynomial with real/complex coefficients and argument $x \in [0,1]$. – madlin Nov 5 '14 at 0:45
• I see. So I guess an algebraic polynomial is the same as a polynomial. : ) – Rudy the Reindeer Nov 5 '14 at 8:25
• Well, yeah, it's in contrast to trigonometric ;) – madlin Nov 5 '14 at 8:58

Polynomials are continuous. It is a theorem in real analysis that the uniform limit of continuous functions is continuous. As a consequence, the closure in the $\sup$-norm of the algebra of polynomials is contained in the set of continuous functions.
Therefore, any bounded function $f$ that is not continuous should convince you that $f$ is not the uniform limit of polynomials.