Two Pathological Examples in Functional Analysis

I have not written here for a long time, and I just realized that I never published this exercise. It was from a class in functional analysis. Its purpose was to have us explicitly demonstrate the existence of a normed vector space that is not complete, i.e., not a Banach space, and of an invertible, bounded, linear operator whose inverse is not bounded.

Equip $\mathbb{N}$ with the discrete topology, let $c_c:=(C_c(\mathbb{N}),\Vert \cdot {\Vert }_{sup})$, i.e., the normed, linear space over $\mathbb{C}$ of continuous, compactly supported functions $\mathbb{N}\to \mathbb{C}$, and note that, since $f\in c_c$ is eventually zero,$$\Vert f{\Vert }_{sup}=\underset{l\in \mathbb{N}}{max}\{|f(l)|\}.$$For every $k\in \mathbb{N}$, define$$\begin{array}{rl}{f}_k:\mathbb{N}& \to \mathbb{C}\text{by}\\ l& \mapsto \{\begin{array}{ll}{l}^{-1}& \text{ if }l\le k\text{ and}\\ 0& \text{ otherwise},\end{array}\end{array}$$note that $f_k\in c_c$, let $\epsilon >0$, let $N>{\epsilon }^{-1}$, let $m,n\in \mathbb{N}$, and assume, without loss of generality, that $m>n>N$. Then$$\Vert f_m-f_n{\Vert }_{sup}=m^{-1}<N^{-1}<\epsilon ,$$i.e., $(f_k{)}_{k=1}^{\mathrm{\infty }}$ is Cauchy.

Define $f:\mathbb{N}\to \mathbb{C}$ by $l\mapsto l^{-1}$, note that $f\notin c_c$, let $\epsilon >0$, let $N>{\epsilon }^{-1}-1$, and let $k\in \mathbb{N}$ such that $k>N$. Then$$\Vert f_k-f{\Vert }_{sup}=(k+1{)}^{-1}<(N+1{)}^{-1}<\epsilon ,$$i.e., $(f_k{)}_{k=1}^{\mathrm{\infty }}$ converges to $f$.

Therefore, $c_c$ is not complete.

Define $$\begin{array}{rl}T:c_c& \to c_c\text{by}\\ f=(f_1,f_2,f_3,\dots )& \mapsto (\frac{f_1}{1},\frac{f_2}{2},\frac{f_3}{3},\dots ).\end{array}$$

Let $f,g\in c_c$ and let $\lambda \in \mathbb{C}$. Then$$\begin{array}{rl}T(\lambda f+g)& =(\frac{\lambda f_1+g_1}{1},\frac{\lambda f_2+g_2}{2},\frac{\lambda f_3+g_3}{3},\dots )\\ & =(\lambda \frac{f_1}{1}+\frac{g_1}{1},\lambda \frac{f_2}{2}+\frac{g_2}{2},\lambda \frac{f_3}{3}+\frac{g_3}{3},\dots )\\ & =\lambda (\frac{f_1}{1},\frac{f_2}{2},\frac{f_3}{3},\dots )+(\frac{g_1}{1},\frac{g_2}{2},\frac{g_3}{3},\dots )\\ & =\phantom{\frac{1}{1}}\lambda T(f)+T(g).\end{array}$$

Let $f=(f_1,f_2,\dots )\in c_c$ and note that$$\Vert T(f){\Vert }_{sup}=\underset{l\in \mathbb{N}}{max}\{\frac{|f_l|}{l}\}\le \underset{l\in \mathbb{N}}{max}\{\frac{\Vert f{\Vert }_{sup}}{l}\}=\Vert f{\Vert }_{sup}.$$

Since $T$ is linear, $T$ is injective.

Let $f=(f_1,f_2,f_3,\dots )\in c_c$ and $g=(1f_1,2f_2,3f_3,\dots )\in c_c$. Then $T(g)=f$.

Define $S:c_c\to c_c$ by $f=(f_1,f_2,f_3,\dots )\mapsto (1f_1,2f_2,3f_3,\dots )$. Then $S\circ T(f)=T\circ S(f)=f$ so that $S=T^{-1}$.

Suppose that $M$ is such that $\Vert T^{-1}(f){\Vert }_{sup}\le M\Vert f{\Vert }_{sup}$ for every $f\in c_c$, let $k\in \mathbb{N}$ such that $k>M$, and define $g:\mathbb{N}\to \mathbb{C}$ by $l\mapsto 1$ if $l=k$ and $l\mapsto 0$ otherwise. Then$$\Vert T^{-1}(g){\Vert }_{sup}=\underset{l\in \mathbb{N}}{max}\{|l{g}_l|\}=k=k\Vert g{\Vert }_{sup}>M\Vert g{\Vert }_{sup},$$which is a contradiction.