Exponential function and uniform convergence of polynomials. How can I prove that no sequence of polynomials converges uniformly to the exponential function?
Thanks in advance for any help.
 A: 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 $|f_n(2)-e^2|>e^2-2>1$, contradiction.
A: 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 $x$.
A: 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.
