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Let $f \in L^1[0,1]$, can we approximate $f$ by a polynomial, in sup norm ?

I know that the algebra of polynomials is dense in algebra of continuous functions, wrt to sup norm, And I know that if $f \in L^1[0,1]$, for any $\epsilon > 0$, we can find a continuous function $g \in C[0,1]$, such that $||f-g||_1=\int^1_0 |f-g| < \epsilon$. these being said, Is the following possible ?

if $f \in L^1[0,1]$, for any $\epsilon > 0$, we can find a polynomial $P$, such that $$\sup_{x\in [0,1]}|f-P| < \epsilon$$ I already appreciate your help.

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    $\begingroup$ you can take any discontinuous function $f\in L_1,$ say $I_{[0,1/2]}$ to get a contradiction. Indeed, if such polynomials exist, the limiting function has to be continuous $\endgroup$ – leshik Feb 27 '14 at 6:46
  • $\begingroup$ Indeed, even for polynomial approximations of that function, as they are used in filter design, one has a region where $|p(0.5+a)-p(0.5-a)|<ϵ$ for some $a>0$, but $|f(0.5+a)-f(0.5-a)|=1$, so that $$|f(0.5+a)-p(0.5+a)|+|f(0.5-a)-p(0.5-a)|\ge 1-ϵ$$ is not smaller than $2ϵ$ for $ϵ<1/3$. $\endgroup$ – Dr. Lutz Lehmann Feb 27 '14 at 6:57
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The function $$ f(x) = \begin{cases} 1, &\text{if $x$ is rational}, \\ 0, &\text{if $x$ is irrational} \end{cases} $$ is measurable and bounded, hence integrable on any bounded interval. But $\sup_{x\in[0,1]} |f(x)-P(x)| \ge \frac12$ for any polynomial $P$ (indeed, any continuous function $P$).

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