 # Let f(x)=\sin(2x). Prove or disprove there are a sequnce of polynomials P_n(x) which convenes to f(x) uniformly on (0,\infty). vazelinahS 2021-09-04 Answered
Let $f\left(x\right)=\mathrm{sin}\left(2x\right)$
Prove or disprove there are a sequnce of polynomials ${P}_{n}\left(x\right)$ which convenes to $f\left(x\right)$ uniformly on $\left(0,\mathrm{\infty }\right)$.
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We have to prove or disprove there are a sequence of polynomials ${P}_{n}\left(x\right)$ which converges to f(x) uniformly on $\left(0,\mathrm{\infty }\right)$
We have to prove by contradiction.
Suppose there exists a sequence of polynomial ${f}_{n}\left(x\right)$ which converges uniformly to $\mathrm{sin}\left(2x\right)$
So $Sup|{f}_{n}\left(x\right)-\mathrm{sin}\left(2x\right)|<ϵ\mathrm{\forall }n\ge {n}_{0}$
In particular for $ϵ=\frac{1}{2}$
We have
$Sup|{f}_{n}\left(x\right)-\mathrm{sin}\left(2x\right)|<\frac{1}{2}$
As $\mathrm{sin}\left(2x\right)$ is bounded function by 1(By triangle inequality)
So, $|{f}_{n}\left(x\right)|<\frac{3}{2}$
This means, sequence of polynomial ${f}_{n}$ is bounded by $\frac{3}{2}$
That is almost all ${f}_{n}$ are constant polynomials
But ${f}_{n}\left(0\right)\to \mathrm{sin}\left(2\cdot 0\right)=0$ as $n\to \mathrm{\infty }$ that is ${f}_{n}\left(0\right)\to 0$
And ${f}_{n}\left(\frac{\pi }{4}\right)\to \mathrm{sin}\left(2\cdot \frac{p}{4}\right)=1$ as $n\to \mathrm{\infty }$ that is ${f}_{n}\left(\frac{\pi }{4}\right)\to 1$
Which leads to contradiction to fact that almost all ${f}_{n}$ are constant polynomial.
Hence, there doesn't exists any sequence of polynomial converges uniformly to $\mathrm{sin}\left(2x\right)$