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Show that there doesn't exist any sequence of polynomials which converge to $\sin x$ uniformly on $\mathbb{R}$.

Suppose there is a sequence of polynomials $\{p_n\}$ which converges to $\sin x$ uniformly on $\mathbb{R}$. Then given $\epsilon \gt 0$, there exists a $n_0 \in \mathbb{N}$ such that $\forall n \ge n_0, |p_n(x)-\sin x|\lt \epsilon.$ In particular for $\epsilon=1$, $|p_n(x)|\lt 2$ for all $n \ge n_0$.This makes $p_{n_0}(x)$ a constant polynomial say $c=p_{n_0}(x)$.

Then I need to find a contradiction somehow.

How do I do that??

Thanks for the help!!

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    $\begingroup$ Since you already know that $p_n$ would be constant for all large enough $n$, $p_n \equiv c_n$, all you need is that a sequence of constant functions cannot converge uniformly to $\sin$. Try $\epsilon = \frac{1}{2}$ for example to prove that. $\endgroup$ – Daniel Fischer Nov 10 '14 at 12:55
  • $\begingroup$ @DanielFischer There is another problem where I need to do the same for $e^{x}$. In this case I am getting a contradiction when I am restricting $e^{x}$ to $x \lt 0$ in place of whole of $\mathbb{R}$. But does that prove my claim? $\endgroup$ – tattwamasi amrutam Nov 10 '14 at 13:11
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    $\begingroup$ If you show that no sequence of polynomials can converge to $f(x)$ uniformly on a subset $A$ of $\mathbb{R}$, then it follows a fortiori that no sequence of polynomials can converge uniformly to $f$ on $\mathbb{R}$. For uniform convergence on a set $S$ implies uniform convergence on all subsets $T\subset S$. $\endgroup$ – Daniel Fischer Nov 10 '14 at 13:17
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As you have shown, almost all $p_n$ are constant. Then from $p_n(0)\to \sin 0=0$ and $p_n(0)=p_n(\pi/2)\to\sin(\pi/2)=1$, you have your contradiction.

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