I know that if $P_n$ are continuous functions and $P_n \rightrightarrows P$, $P$ is also continuous function. But I can't see in which direction I should dig to prove that $P$ is polynomial.

I will appreciate any hint and help.

  • $\begingroup$ What does the double arrow mean? $\endgroup$ – Francesco Polizzi Apr 19 at 18:57
  • $\begingroup$ uniform convergence $\endgroup$ – mierzej Apr 19 at 19:02
  • $\begingroup$ Since bounded polynomials are constant, the uniform Cauchy criterion shows that there is $N \in \mathbb{Z}^+$ such that for all $n \geq N$, the function $C_n := P_n-P_N$ is constant (as a function of $x$). On the other hand $C_n$ converges to $C = P-P_N$, which is constant, so $P = P_N + C$. $\endgroup$ – Pete L. Clark Apr 19 at 19:28
  • $\begingroup$ (This level of question is probably more appropriate for math.stackexchange.com though.) $\endgroup$ – Pete L. Clark Apr 19 at 19:31
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    $\begingroup$ @mierzej MathOverflow is for research-level mathematics, while MathStackExchange is for general mathematics questions. $\endgroup$ – Antoine Labelle Apr 19 at 20:08

We have $(P_n-P_m)\rightrightarrows 0$ if $n,m$ tend to infinity. Since $P_n-P_m$ is a polynomial, this yields that degrees of $P_n$ are uniformly bounded, say they do not exceed $d$. Now even the pointwise convergence in $d+1$ points yields the coefficientwise convergence (by Lagrange interpolation, for example), hence on any segment $P_n\rightrightarrows P_0$ where $P_0$ is this limit polynomial. Hence $P\equiv P_0$.

  • $\begingroup$ When you say "Now even the pointwise convergence in $d+1$ points yields the coefficientwise convergence" you mean that every coefficient in $P_n$ is is convergent to one of $P$ coefficients? $\endgroup$ – mierzej Apr 19 at 19:17
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    $\begingroup$ Yes, if $P_n=\sum_{k=0}^d a_{kn}x^k$, then $a_{kn}$ has a limit $a_k$ when $n$ goes to infinity. Then we define $P_0=\sum_{k=0}^d a_kx^k$. $\endgroup$ – Fedor Petrov Apr 19 at 19:20
  • $\begingroup$ oooo I see, thanks a lot $\endgroup$ – mierzej Apr 19 at 19:22

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