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.