Let $\{p_k\} \subset P_n \subset C_2[a,b]$ and suppose $\|p_k\| \rightarrow 0$. Show that $p_k \rightarrow 0$ uniformly on $[a,b]$.

Question

Task: Let $\{p_k\} \subset P_n \subset C_2[a,b]$ and suppose $\|p_k\| \rightarrow 0$. Show that $p_k \rightarrow 0$ uniformly on $[a,b]$. That is, for each $\epsilon > 0$, there exists an integer $N$ such that $\sup_{x \in [a,b]} |p_k(x)| < \epsilon$ whenever $k \geq N$.

My attempt: I was given a hint to use the next theorem here: Let $M$ be a finite-dimensional subspace of the inner product space $X$. Suppose $\{x_1, x_2, \ldots, x_n\}$ is a basis for $M$, $y_k = \sum_{i=1}^n \alpha_{ki} x_i$, and $y = \sum_{i=1}^n \alpha_i x_i$. Then $y_k \rightarrow y$ if and only if $\alpha_{ki} \rightarrow \alpha_i$ for each $i = 1, 2, \ldots, n$, but I just don't even know how to start. Any idea?

Note that he set of all polynomials of degree at most $n$,

$$ P_n = \left\{ x \ \Bigg| \ x(t) = \sum_{i = 0}^n \alpha_i t^i, \ \alpha_i \in \mathbb{R} \right\} $$

is an $(n+1)$-dimensional subspace of $C_2[a, b]$. Also: $\|f\| = (\int_{a}^{b} |f(x)|^2 \,dx)^\frac{1}{2}$ $\forall f \in C_2[a, b]$.

Answer 1

For each $f,g\in P_n$ put $(f,g)=\int_a^b f(x)g(x)dx$. It is easy to check that $(\cdot,\cdot)$ is an inner product on $P_n$. Let $\{x^0|[a,b],x^1|[a,b],\dots,x^n|[a,b]\}$ be the basis of $P_n$. For each natural $k$ let $p_k=\sum_{i=0}^n \alpha_{ki} x^i|[a,b]$ and let $0=\sum_{i=0}^n 0 x^i|[a,b]\equiv \sum_{i=1}^n \alpha_{i} x^i|[a,b]$. The theorem from the hint implies that for each nonnegative integer $i\le n$ we have $\lim_{k\to\infty} \alpha_{ki}=\alpha_i=0$. Given any $\epsilon>0$ pick $\delta>0$ such that $(n+1)(|a|+|b|+1)^n\delta<\epsilon$. Pick natural $N$ such that for each nonnegative integer $i\le n$ and for each natural $k>N$ we have $|\alpha_{ki}|<\delta$. Then $\|p_k\|<\epsilon$.

Answer 2

The $n+1$-dimensional subspace $P_n$ of the inner product space $C^2[a,b]$ is also an inner product space.

Define $I_k:P_N \to {\mathbb R}$ by letting $p = \sum_{k=0}^n a_k x^k$, let $I_k(p) = a_k$. Then $I_k$ is linear mapping from an $n+1$-dimensional space to a $1$ dimensional space, and therefore can be represented through multiplication by $1\times n$ matrix, which is equivalent to taking the inner product with some element in $q_k$ in $P_n$ (to see, obtain an ON basis from Gram-Schmidt first). In other words, there exists a unique $q_k\in P_n$ such that

$$I_k(p) = \int p q_k\mbox{ for all }p\in P_n.$$

Apply Cauchy-Schwarz to conclude:

$$|I_k(p)|\le \|q_k\|\|p\|.$$

Let $(p_\ell:\ell\in{\mathbb N})$ be a sequence in $P_n$ converging to $0$ in norm. By construction, $p_\ell = \sum_{k=0}^n I_k(p_\ell) x^k$. Then the last inequality gives

$$\lim_{\ell\to\infty} I_k(p_\ell)=0\mbox{ for all }k\in \{0,\dots,n\}.$$ Now,

$$|p_\ell(x)|\le \sum_{k=0}^n |I_k(p_\ell)||b|^k\le (1+|b|)^n \sum_{k=0}^n |I_k(p_\ell)|,$$ where for the first inequality we assumed without loss of generality $|b|\ge|a|$, and the second inequality follows from $|b|^k\le (1+ |b|)^k \le (1+|b|)^n$.

To conclude,

$$ \sup_{x\in[a,b]} |p_\ell (x)|\le (1+|b|)^n \sum_{k=0}^n |I_k(p_\ell)|\to 0.$$