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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?

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  • $\begingroup$ What is an algebraic polynomial? $\endgroup$ Nov 5, 2014 at 0:37
  • $\begingroup$ It is a polynomial with real/complex coefficients and argument $x \in [0,1]$. $\endgroup$
    – madlin
    Nov 5, 2014 at 0:45
  • $\begingroup$ I see. So I guess an algebraic polynomial is the same as a polynomial. : ) $\endgroup$ Nov 5, 2014 at 8:25
  • $\begingroup$ Well, yeah, it's in contrast to trigonometric ;) $\endgroup$
    – madlin
    Nov 5, 2014 at 8:58

1 Answer 1

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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.

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