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## Approximation by polynomials of a function defined on the entire real line

Suppose ${f: \Bbb{R}\rightarrow \Bbb{R}}$ is a function which can be uniformly approximated by polynomials on ${\Bbb{R}}$. Then ${f}$ is also a polynomial.

Note the striking difference between this result and the Weierstrass approximation theorem.

The proof is very simple, and it relies on the basic properties of polynomials. Suppose that ${f}$ can be approximated uniformly by a sequence ${(P_n)}$ of polynomials on the entire real line. Then for ${n}$ great enough and ${k\geq 1}$ we have

$\displaystyle |P_n(x)-P_{n+k}(x)|\leq |P_n(x)-f(x)|+|P_{n+k}(x)-f(x)|<2\varepsilon$

for every ${k \geq 1}$. But the difference of two polynomials is again a polynomial, and every nonconstant polynomial is unbounded on ${\Bbb{R}}$. This means that for ${n}$ great enough the polynomials are essentially the same. The only thing that can differ is the free coefficient ${c_n}$. But since ${P_n(0)=c_n}$ is convergent to ${f(0)}$ it follows that ${c_n}$ is also convergent, and trivially ${P_n}$ uniformly converges to ${P_n(x)-P(0)+f(0)}$.

Since the uniform limit is unique it follows that ${f}$ is a polynomial.

There is a similar result presented here. If a function ${g}$ defined on a compact interval ${[a,b]}$ is uniformly approximated by polynomials whose degrees are bounded above, then ${g}$ is also a polynomial.