Pointwise convergence of polynomials of degree $k$

165 Views Asked by At

Let $\{P_ n\}$ be a sequence of real polynomials with highest degree $k$. I know that if $P_n$ converges uniformly on $[0,1]$, then it must converge to a polynomial. I am curious that if it is possible that $P_n$ converges pointwise to a non-polynomial and nonconstant function on $[0,1]$. Can anyone give some examples?

2

There are 2 best solutions below

0
On BEST ANSWER

If degrees of polynomials are bounded (say by $k$) then these polynomials can be viewed as vectors in final dimensional space spanned by ${1,x,x^2,...,x^k}$. Since in finite dimensional space unit balls are compact (hence any ball with finite radius are compact) every bounded sequence contains convergent sub-sequence. Hence, the limit belongs to the space i.e. it is a polynomial. However, if degrees are not bounded then of course you can have limit which is not a polynomial, say $$p_n(x) = \sum_{k=0}^n \frac{x^k}{k!}$$ converges uniformly to $e^x$ on $[0,1]$.

0
On

In fact, it is only necessary that the polynomials converge pointwise at $k + 1$ distinct points.

Indeed, suppose we have a set of $k + 1$ distinct points $a_0, \ldots, a_k$, and for all $i$, $\lim\limits_{n \to \infty} P_n(a_i)$ exists.

Let $Q_i$ be the polynomial such that $Q_i(a_j) = \delta_{ij}$. These $k + 1$ polynomials are linearly independent; therefore, they form a basis. In particular, the coefficient of $Q_i$ in $R$ is $R(a_i)$. Thus, the fact that $P_n(a_i)$ converges means that each of the coefficients of $Q_i$ converge. So the polynomials thus converge uniformly to $\sum\limits_{i = 0}^k Q_i \lim\limits_{n \to \infty} P_n(a_i)$, which is a polynomial of degree at most $k$.