# If $P_n \rightrightarrows P$ in $\mathbb{R}$ and $P_n$ are polynomials proof that $P$ is polynomial [closed]

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.

• What does the double arrow mean? – Francesco Polizzi Apr 19 at 18:57
• uniform convergence – mierzej Apr 19 at 19:02
• 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 $P = P_N + C$. – Pete L. Clark Apr 19 at 19:28
• (This level of question is probably more appropriate for math.stackexchange.com though.) – Pete L. Clark Apr 19 at 19:31
• @mierzej MathOverflow is for research-level mathematics, while MathStackExchange is for general mathematics questions. – 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 $$P\equiv P_0$$.

• 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? – mierzej Apr 19 at 19:17
• 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 $P_0=\sum_{k=0}^d a_kx^k$. – Fedor Petrov Apr 19 at 19:20
• oooo I see, thanks a lot – mierzej Apr 19 at 19:22