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Let $f$ is continuous function on $[0,1]$ and $\epsilon>0$. Show that there is a polynomial $p$ such that $p(0)=f(0)$, $p(1)=f(1)$, $f(x)-p(x)|<\epsilon, \forall x\in(0,1)$.

Consider $f(x)-p(x)=x(x-1)q(x)$. Clearly it satisfies the given condition. Then $|f(x)-p(x)|=|x(x-1)q(x)|.$

Please help me to complete the proof.

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  • This is Weierstrass Approximation Theorem. A proof can be found https://math.stackexchange.com/questions/4100013/proving-weierstrass-approximation-theorem. – Pleroma Feb 22 '22 at 07:07
  • @Pleroma not entirely true. The WAT may not have the boundary conditions satisfied. – Just a user Feb 22 '22 at 08:48
  • @Justauser You're right. This is not exactly the WAT, but the approximate polynomials constructed by Bernstein polynomials satisfy the boundary conditions. – Pleroma Feb 22 '22 at 09:35
  • @Pleroma I see, thanks for telling me! – Just a user Feb 22 '22 at 09:51

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