Since a Banach space is a complete normed vector space and a norm always induces a metric, a Banach space must be a metric space, right?
If so, why is a Banach space defined as a complete normed space and not a complete metric space?
[Math] Is a Banach space also a metric space
banach-spacesmetric-spacesnormed-spaces
Related Solutions
By definition $X'$ is a space of bounded linear functionals on $X$. More preciesly $$ X'=\{f\in\mathcal{L}(X,\mathbb{C}):\Vert f\Vert<+\infty\} $$ where $\mathcal{L}(X,\mathbb{C})$ is a linear space of linear functions from $X$ to $\mathbb{C}$ and $$ \Vert f\Vert:=\sup\{|f(x)|:x\in X\quad \Vert x\Vert\leq 1\} $$ In order to prove that $X'$ is complete consider Cauchy sequence $\{f_n:n\in\mathbb{N}\}\subset X'$. Fix $\varepsilon>0$. Since $\{f_n:n\in\mathbb{N}\}$ is a Cauchy sequence there exist $N\in\mathbb{N}$ such that for all $n,m>N$ we have $\Vert f_n-f_m\Vert\leq\varepsilon$. Consider arbitrary $x\in X$, then $$ |f_n(x)-f_m(x)|=|(f_n-f_m)(x)|\leq\Vert f_n-f_m\Vert\Vert x\Vert\leq\varepsilon \Vert x\Vert $$ Thus we see that $\{|f_n(x)|:n\in\mathbb{N}\}\subset\mathbb{C}$ is a Cauchy sequence. Since $\mathbb{C}$ is complete, there exist unique $\lim\limits_{n\to\infty}f_n(x)$. Since $x\in X$ is arbitrary we can define function $$ f(x):=\lim\limits_{n\to\infty}f_n(x) $$ Our aim is to show that $f\in X'$ and $\lim\limits_{n\to\infty}f_n=f$. Let $x_1,x_2\in X$, $\alpha_1,\alpha_2\in\mathbb{C}$ then $$ f(\alpha_1 x_1+ \alpha_2 x_2)= \lim\limits_{n\to\infty}f_n(\alpha_1 x_1+ \alpha_2 x_2)= $$ $$ \lim\limits_{n\to\infty}(\alpha_1 f_n(x_1) + \alpha_2 f_n(x_2))= \alpha_1 \lim\limits_{n\to\infty}f_n(x_1) + \alpha_2 \lim\limits_{n\to\infty}f_n(x_2))= \alpha_1 f(x_1) + \alpha_2 f(x_2) $$ So we conclude $f\in\mathcal{L}(X,\mathbb{C})$. Since $\{f_n:n\in\mathbb{N}\}$ is a Cauchy sequence it is bounded in $X'$, i.e. there exist $C>0$ such that $\sup\{\Vert f_n\Vert:n\in\mathbb{N}\}\leq C$. Hence, for all $x\in X$ we have $$ |f(x)|= |\lim\limits_{n\to\infty}f_n(x)|= \lim\limits_{n\to\infty}|f_n(x)|\leq \limsup\limits_{n\to\infty}\Vert f_n\Vert \Vert x\Vert\leq \Vert x\Vert\sup\{\Vert f_n\Vert:n\in\mathbb{N}\}\leq C\Vert x\Vert $$ Now we see that $\Vert f\Vert\leq C$, but as we proved earlier $f\in\mathcal{L}(X,\mathbb{C})$, so $f\in X'$. Finally recall that for given $\varepsilon>0$ and $x\in X$ there exist $N\in\mathbb{N}$ such that $n,m>N$ implies $$ |f_n(x)-f_m(x)|\leq\varepsilon \Vert x\Vert. $$ Then let's take here a limit when $m\to\infty$. We will get $$ |f_n(x)-f(x)|\leq\varepsilon \Vert x\Vert. $$ Since $x\in X$ is arbitrary we proved that for all $\varepsilon>0$ there exist $N\in\mathbb{N}$ such that $n>N$ implies $$ \Vert f_n-f\Vert= \sup\{|f_n(x)-f(x)|:x\in X,\quad \Vert x\Vert\leq 1\}\leq \varepsilon. $$ This means that $\lim\limits_{n\to\infty} f_n=f$. Since we showed that every Cauchy sequence in $X'$ have a limit, $X'$ is complete.
This proof can be easily generalized up to the following theorem: If $X$ is a normed space and $Y$ is a Banach space, then the linear space of all bounded linear functions from $X$ to $Y$ is complete.
Hints: 1)There is no norm on on $C^{\infty}[0,1]$ which makes the derivative continuous. This is because, $||\frac{d}{dx}(e^{nx})|| \leq c||e^{nx}||$ for all values of $n$ isn't possible.
2) Show that in the above metric the derivative map is continuous.
Best Answer
Every Banach space is a metric space. However, there are metrics that aren't induced by norms. If this were the case, then Banach spaces and complete metric spaces would the same thing... if metric spaces had operations and an underlying field! The difference is more than just the metric/norm dichotomy. For completeness of the answer (no pun intended), a metric $d:X\times X \to \Bbb R_{\geq 0}$ comes from a norm if and only if: $$d(x+z,y+z) = d(x,y)\quad\text{and}\quad d(\lambda x,\lambda y) = |\lambda|d(x,y),$$ for all $x,y,z \in X$, for all $\lambda \in \Bbb R$. If this is the case, the norm is $\|x\| = d(x,0)$.