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.
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$\begingroup$ What does the double arrow mean? $\endgroup$– Francesco PolizziCommented Apr 19, 2021 at 18:57
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$\begingroup$ uniform convergence $\endgroup$– mierzejCommented Apr 19, 2021 at 19:02
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$\begingroup$ 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. ClarkCommented Apr 19, 2021 at 19:28
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$\begingroup$ (This level of question is probably more appropriate for math.stackexchange.com though.) $\endgroup$– Pete L. ClarkCommented Apr 19, 2021 at 19:31
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2$\begingroup$ @mierzej MathOverflow is for research-level mathematics, while MathStackExchange is for general mathematics questions. $\endgroup$– Antoine LabelleCommented Apr 19, 2021 at 20:08
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1 Answer
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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 $P\equiv P_0$.
answered Apr 19, 2021 at 19:09
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$\begingroup$ 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? $\endgroup$– mierzejCommented Apr 19, 2021 at 19:17
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1$\begingroup$ 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 $P_0=\sum_{k=0}^d a_kx^k$. $\endgroup$ Commented Apr 19, 2021 at 19:20
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