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How can I prove that no sequence of polynomials converges uniformly to the exponential function?

Thanks in advance for any help.

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  • $\begingroup$ The exponential function grows too fast. $\endgroup$ – Angela Richardson Jun 12 '13 at 17:55
  • $\begingroup$ Yes, I tried doing it by assuming that there was a uniformly convergent sequence of polynomial to see if I got a contradiction but I'm not really sure where to look. $\endgroup$ – LaurentP Jun 12 '13 at 18:07
  • $\begingroup$ Do you know "Weierstrass approximation theorem"? You should have a look at it. $\endgroup$ – Mhenni Benghorbal Jun 12 '13 at 18:25
  • $\begingroup$ @AngelaRichardson It also decays too fast without being zero, cf. my proof that avoids using the "big end" on the right. $\endgroup$ – Hagen von Eitzen Jun 12 '13 at 18:28
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If a sequence $(f_n)$ of polynomials converges uniformly to $x\mapsto e^x$ on all of $\mathbb R$ (or, as the following argument merely assumes, just on $(-\infty,2]$), then for $\epsilon=1$ there exists $N$ such that $|f_n(x)-e^x|<1$ for all $n>N$ and all $x\in\mathbb R$. Since $|e^x|<1$ for $x<0$, we conclude that $|f_n(x)|<2$ for $x<0$. It is easy to see that this is the case only for constant polnomials (all others go to $\pm\infty$ as $x\to-\infty$) and so $f_n(x)=c$ with $|c|<2$. But then $|f(2)-e^2|>e^2-2>1$, contradiction.

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Suppose that $f:\mathbb R\to\mathbb R$ is a uniform limit of polynomial functions. Then $f$ is a polynomial function.

Proof Sketch: Suppose $(p_n)$ is a sequence of polynomial functions converging uniformly to $f$. Then there is a positive integer $N$ such that $m,n\geq N$ implies $|p_n(x)-p_m(x)|<1$ for all $x\in\mathbb R$. That means that the polynomial function $p_n-p_m$ is bounded, hence constant. Thus our sequence of polynomials has the form $(p_1,p_2,\ldots,p_N,p_N+a_1,p_N+a_2,p_N+a_3,\ldots)$ for some sequence $(a_n)$ of real numbers. For the limit to exist it must be the case that $(a_n)$ converges to some real number $a$, and $f(x)=p_N(x)+a$ for all $x$.

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If such a sequence of polynomials existed, then for any $\epsilon$, there would be $N$ such that for any $x\in\mathbb{R}$ we have

$$|p_n(x)-e^x|<\epsilon$$

so long as $n>N$. However, we have $p_{N+1}(x)<e^x-1$ for large enough $x$, so that $|p_{N+1}-e^x|>1$, a contradiction.

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