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Suppose $p_n(x)$ is a sequence of polynomials which converge to a function $f$, uniformly on $\mathbb{R}$. Show that $f$ is a polynomial.

If there were a uniform bound $M$ on the degree of $p_n(x)$, then by differentiating term by term, we could show that $\partial_x^M f \equiv 0$, and $f$ is a polynomial. But in the absence of such a bound, I'm not sure...

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  • $\begingroup$ The power series of $f$ might be written as a polynomial with the given polynomials as a basis. However, this might be a specific case. $\endgroup$ Jul 27, 2013 at 21:59
  • $\begingroup$ Does it help to take a subsequence for which the degree is monotone increasing? $\endgroup$
    – dfeuer
    Jul 27, 2013 at 22:25

1 Answer 1

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The sequence $(p_n)$ satisfies the uniform Cauchy criterion; so one can find $n_0$ such that for all $n\geq n_0$, we have $\sup_{x\in\mathbb R} \vert p_n(x)-p_{n_0}(x)\vert\leq 1$. Now, a polynomial function is bounded on $\mathbb R$ only if it is constant. So for $n\geq n_0$ we have $p_n(x)=p_{n_0}(x)+c_n$ for some constant $c_n$. Since $p_n\to f$ pointwise, the sequence $(c_n)$ is convergent with limit $c=f(0)-p_{n_0}(0)$ and we have $f(x)=p_{n_0}(x)+c$ for all $x\in\mathbb R$.

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  • $\begingroup$ Thank you! I understand now. $\endgroup$
    – Eric Auld
    Jul 27, 2013 at 22:26

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