This problem was given as an exercise in a Functional Analysis lecture. I’ve tidied up my old solution for the problem, and here it is.

Definitions

Definition 1. (Sub-vector Space)

Suppose that \(V\) and \(W\) are two vector spaces and \(0_V \in W \subseteq V\). W is a sub-vector space of V iff $$\forall w_1, w_2 \in W \land v \in V : v(w_1 + w_2) \in W$$

Definition 2. (Metric Space)

Let \(\empty \neq X\) be a set and \(d : X \times X \mapsto \R\) be a function. The function \(d\) is called a metric function on \(X\) if \(d\) satisfies

  1. \(d(x, x) = 0\) and \(d(x, y) > 0\) for \(x \neq y\),
  2. \(d(x, y) = d(y, x)\),
  3. \(d(x, z) \leq d(x, y) + d(y, z)\)

for all \(x, y, z \in X\). The tuple \((X, d)\) is called a metric space.

Definition 3. (Cauchy Series)

Let \((X, d)\) be a metric space and \((x_n) \subset X\) be a sequence. If there exists an \(N_\epsilon \in \N\) such that \(d(x_n, x_m) < \epsilon\) for an arbitrary \(\epsilon > 0\) when \(n, m > N_\epsilon \in \N\), then the sequence \((x_n)\) is called a Cauchy sequence in \(X\).

The sequence \((x_n)\) being a Cauchy sequnce is equivalent to \(\lim\limits_{n, m \to \infty} d(x_n, x_m) = 0\).

Definition 4. (Complete Metric Space)

Let \((X, d)\) be a metric space. If all Cauchy series in \(X\) converges to a point in \(X\), in other words, if \(\lim\limits_{x \to \infty} x_n = x \in X\) for all \((x_n) \subset X\), then \((X, d)\) is a complete metric space.

Problem

Let \(\mathcal{P}([a, b])\) be the set of real valued polynomial functions defined on the closed interval \([a, b]\) and \(\mathcal{C}([a, b])\) be the set of real valued continuous functions defined on the closed interval \([a, b]\).

  • a) Show that \(\mathcal{P}([a, b])\) is a subvector space of \(\mathcal{C}([a, b])\).

For some \(n, m \in \N \cup \{0\}\), let \((a_k)_{k=0}^{\infty} \subset \R , \forall i > n : a_i = 0\), \((b_k)_{k=0}^{\infty} \subset \R , \forall i > m : b_i = 0\), \(f(x) = \sum_{i = 0}^{\infty} a_ix^i\), \(g(x) = \sum_{i = 0}^{\infty} b_ix^i\) for arbitrary real polynomial functions \(f, g \in \mathcal{P}([a, b])\).

\(a_j + b_j \in \R\) for all \(j \in \{0, 1, 2, \cdots\}\). Hence,

$$\begin{aligned}f(x) + g(x) &= \sum_{i=0}^{\infty} a_ix^i + \sum_{i=0}^{\infty} b_ix^i = \sum_{i=0}^{\infty} (a_i + b_i)x^i \\ &= (a_0 + b_0) + \cdots + (a_n + b_n)x^n + \cdots + (a_m + b_m)x^m \\ &= (f + g)(x) \in \mathcal{P}([a, b]) \end{aligned}$$

\(\alpha a_j \in \R \) for all \(\alpha \in \R\) and \(j \in \{0, 1, 2, \cdots\}\). Hence,

$$\begin{aligned}\alpha f(x) &= \alpha\sum_{i=0}^{\infty} a_ix^i = \sum_{i=0}^{\infty} \alpha a_ix^i \\ &= \alpha a_0 + \cdots + \alpha a_nx^n \\ &= (\alpha f)(x) \in \mathcal{P}([a, b]) \end{aligned}$$

Thus, \(\mathcal{P}([a, b])\) is a sub-vector space of \(\mathcal{C}([a, b])\).

  • b) Show that the metric space \((\mathcal{P}([a, b]), d_{\infty})\) is not a complete metric space.

Proof of showing that \( (\mathcal{P}([a, b]), d_{\infty}) \) is a metric space is omitted.

Let \((f_n)_{n=1}^{\infty} = ((1 + \frac{x}{n})^n)_{n=1}^{\infty} \subset \mathcal{P}([a, b])\) be a sequence of functions, $$f: \R \mapsto \R, f(x) = e^x$$ and $$g(x) = |f_n(x) - f(x)| = |(1 + \frac{x}{n})^n - e^x|$$

The function \(f_n(x) = (1 + \frac{x}{n})^n\) is continuous for all \(n \in \N\) on the closed interval \([a, b]\) according to addition and multiplication rule for continuous functions, the absolute value function \(m(x) = |x|\) and the exponential function \(f(x) = e^x\) is continuous on \(\R\) by their definition. Hence, addition, multiplication and chain rule for continuous functions, the function $$g(x) = \left| \left( 1 + \frac{x}{n} \right)^n - e^x \right| = m(f_n(x) - f(x))$$ is continuous on the closed interval \([a, b]\) for all \(n \in \N\).

The closed interval \([a, b] \subset \R\) is a compact set on \(\R\) and g is continuous on the closed interval \([a, b]\) for all \(n \in \N\), hence the Extreme Value Theorem $$\exists x_0 \in [a, b] : g(x_0) = \max\{ g(x) : x \in [a, b] \}$$

So, $$\begin{aligned}d_{\infty}(f_n, f) &= \max\{ | f_n(x) = f(x) | : x \in [a, b] \} \\ &= \max\{ g(x) : x \in [a, b] \} = g(x_0) \end{aligned}$$ for an \(x_0 \in [a, b]\).

From addition rule, multiplication with scalar rule and logaritmic limit method for real limits, $$\begin{aligned}\lim\limits_{n \to \infty} d_{\infty}(f_n, f) &= \lim\limits_{n \to \infty} g(x_0) = \lim\limits_{n \to \infty} \left| \left( 1 + \frac{x_0}{n} \right)^n - e^{x_0} \right| \\ &= \left| \lim\limits_{n \to \infty} \left( 1 + \frac{x_0}{n} \right)^n - \lim\limits_{n \to \infty} e^{x_0} \right| \\ &= |e^{x_0} - e^{x_0}| = |0| = 0 \end{aligned}$$

Thus, the function sequence \( (f_n)_{n=1}^{\infty} \subset \mathcal{P}([a, b]) \) is a convergent sequence (Cauchy sequence), but the convergence point \( f(x) = e^x \notin \mathcal{P}([a, b]) \). So, by the sequence definition of complete metric space, the metric space \( (\mathcal{P}([a, b], d_{\infty})) \) is not a complete metric space.