Let \((P_n)\) be real polynomials whose degrees are bounded by a fixed integer \(D\). Suppose \[ P_n(x)\longrightarrow0 \] for every \(x\in[0,1]\). Prove that \(P_n\to0\) uniformly on \([0,1]\).
Solution
1. Choose \(D+1\) distinct points\[ x_0,x_1,\ldots,x_D\in[0,1] \]and define the Lagrange polynomials\[ L_j(x) \coloneqq \prod_{\substack{0\leq k\leq D\\k\neq j}} \frac{x-x_k}{x_j-x_k}. \]
Such points exist because \([0,1]\) contains infinitely many points. The displayed expression is well defined because the \(x_j\) are pairwise distinct.
2. Every polynomial \(P\) of degree at most \(D\) satisfies\[ P(x) = \sum_{j=0}^{D}P(x_j)L_j(x). \]
The right-hand side is a polynomial of degree at most \(D\) and agrees with \(P\) at each of the \(D+1\) points \(x_0,\ldots,x_D\), because \[ L_j(x_k)= \begin{cases} 1,&j=k,\\ 0,&j\neq k. \end{cases} \] Their difference therefore has degree at most \(D\) and at least \(D+1\) distinct zeros, so it is the zero polynomial.
3. For every \(n\),\[ \norm{P_n}_\infty \leq \sum_{j=0}^{D} \abs{P_n(x_j)}\norm{L_j}_\infty, \]where the sup norms are taken on \([0,1]\).
Apply step 2 to \(P_n\). For each \(x\in[0,1]\), \[ \abs{P_n(x)} \leq \sum_{j=0}^{D} \abs{P_n(x_j)}\abs{L_j(x)}. \] Taking the supremum over \(x\in[0,1]\) gives the displayed inequality. Each \(\norm{L_j}_\infty\) is finite because \(L_j\) is continuous on the compact interval \([0,1]\).
4. The right-hand side in step 3 tends to \(0\) as \(n\to\infty\).
For every fixed \(j\), the assumed pointwise convergence gives \[ P_n(x_j)\longrightarrow0. \] There are only finitely many indices \(j\), and the numbers \(\norm{L_j}_\infty\) are fixed. Hence the finite sum in step 3 tends to \(0\).
5. Therefore\[ \boxed{ \norm{P_n}_\infty\longrightarrow0. } \]Equivalently, \(P_n\to0\) uniformly on \([0,1]\).
Combine steps 3 and 4.
6. Q.E.D.
Step 5 is the required conclusion.