Let $X$ be the vector space of all real sequences . Does there exist a norm on $X$ which makes it complete ?
[Math] Can the real vector space of all real sequences be normed so that it is complete
banach-spacescomplete-spacesfunctional-analysismetric-spacesnormed-spaces
Related Solutions
For future students, here is a more general result:
Let $X$ and $Y$ be normed linear spaces, and let $B(X,Y)$ denote the collection of all bounded linear operators from $X$ to $Y$ endowed with the operator norm. Show that $B(X,Y)$ is a normed linear space, and $B(X,Y)$ is a Banach space whenever $Y$ is a Banach space. The vector operations in $B(X,Y)$ are defined pointwise, i.e. $(A+B)(x)=Ax+Bx$, and $(\alpha A)(x)=\alpha (Ax)$. (Notice that in your case $X'=B(X,\mathbb{C})$ and $\mathbb{C}$ is a Banach space)
It is clear that linear operators form a linear space. To show that $B(X,Y)$ is a linear subspace, it is enough to show the closure to addition and scalar multiplication. But these follow easily from the properties of a norm (the fact that the operator norm satisfies all the properties of a norm for bounded functionals is an easy exercise that follows from properties of supremums in $[0, \infty)$) , namely for any $A,B \in B(X,Y)$ and $\lambda \in \mathbb{C}$ $$\|A+B\| \leq \|A\|+\|B\| < \infty$$ $$\|\lambda A\|=|\lambda| \cdot \|A\| < \infty$$ Thus, $B(X,Y)$ is a normed linear space.
Now assume that $Y$ is a Banach space. Let $\{A_i\}$ be a Cauchy sequence in $B(X,Y)$, i.e. $\forall \, \epsilon >0$, $\exists \, N \in \mathbb{N}$ such that $\forall \, m,n > N$, $\|A_n-A_m\|< \epsilon $. Let $x \in X$ be arbitrary. Let $\epsilon>0$ be arbitrary. If $x=0$, then $$\|A_nx-A_mx\|=0<\epsilon.$$ If $x \neq 0$, choose $N$ such that $\|A_n-A_m\|< \frac{\epsilon}{\|x\|}$. Then by a property of the operator norm, $\forall \, m,n > N$, \begin{equation} \begin{split} \|A_nx-A_mx\| & = \|(A_n-A_m)x\|\\ & \leq \|(A_n-A_m)\| \cdot \|x\|\\ & < \frac{\epsilon}{\|x\|} \cdot \|x\|\\ & = \epsilon\\ \end{split} \end{equation}
Thus, in both cases $\{A_nx\}$ is a Cauchy sequence in $Y$. Since $Y$ is a Banach space, it is convergent to some element in $Y$. Call that element $Ax$, i.e. $$\lim_{n \rightarrow \infty} A_nx=Ax$$ Since $x$ was arbitrary, $Ax$ is defined for any $x \in X$. Thus, $A$ is a map from $X$ to $Y$ defined by $x \rightarrow Ax$. We need to show that $A$ is linear, bounded, and $A_n \xrightarrow{n \rightarrow \infty} A$ in the operator norm. Notice that $A$ is linear, since by linearity of $A_n$ we get that for any $x_1, x_2 \in X$, $\lambda \in \mathbb{C}$, \begin{equation} \begin{split} A(x_1+x_2) & = \lim_{n \rightarrow \infty} A_n(x_1+x_2)\\ & = \lim_{n \rightarrow \infty} (A_nx_1+A_nx_2)\\ & = \lim_{n \rightarrow \infty} A_nx_1+\lim_{n \rightarrow \infty} A_nx_2\\ & = Ax_1+Ax_2\\ \end{split} \end{equation} \begin{equation} \begin{split} A(\lambda x_1) & = \lim_{n \rightarrow \infty} A_n(\lambda x_1)\\ & = \lim_{n \rightarrow \infty} \lambda \cdot A_nx_1\\ & = \lambda \lim_{n \rightarrow \infty} A_nx_1\\ & = \lambda\cdot Ax_1\\ \end{split} \end{equation}
Now recall that Cauchy sequences are bounded. Thus, $\forall \, n$, $\|A_n\|<C$ for some $C \in \mathbb{R}$. Using this fact, we can see that $A$ is bounded, since by continuity of a norm: \begin{equation} \begin{split} \|A\| & =\sup_{\|x\| \leq 1} \|Ax\|\\ & =\sup_{\|x\| \leq 1} \|\lim_{n \rightarrow \infty} A_nx\|\\ & =\sup_{\|x\| \leq 1} \lim_{n \rightarrow \infty} \|A_nx\|\\ & =\sup_{\|x\| \leq 1} \limsup_{n \rightarrow \infty} \|A_nx\|\\ & \leq \sup_{\|x\| \leq 1} \limsup_{n \rightarrow \infty} \Big(\|A_n\|\cdot \|x\|\Big)\\ & \leq \sup_{\|x\| \leq 1} C \cdot \|x\|\\ & = C \sup_{\|x\| \leq 1} \|x\|\\ & \leq C \\ \end{split} \end{equation}
Finally, we want to show that $A_n \xrightarrow{n \rightarrow \infty} A$ in the operator norm. Let $\epsilon > 0$ be arbitrary. Recall that for an arbitrary $x \in X$, we have $$\|A_nx-A_mx\| \leq \|(A_n-A_m)\| \cdot \|x\|$$ Since $\{A_n\}$ is Cauchy, choose $N$ big enough such that for all $n,m \geq N$, $\|(A_n-A_m)\| < \epsilon$. Then the above inequality turns into $$\|A_nx-A_mx\| \leq \epsilon \cdot \|x\|$$ Now by continuity of a norm, we can take limit on both sides as $m$ goes to infinity to obtain $$\|A_nx-Ax\| \leq \epsilon \cdot \|x\|$$ Now taking supremum on both sides over all $x$ such that $\|x\| \leq 1$ yields $$\sup_{\|x\| \leq 1}\|A_nx-Ax\| \leq \epsilon$$ But this is equivalent to saying that for all $n \geq N$, $$\|A_n-A\| \leq \epsilon$$ And since $\epsilon$ was arbitrary, this implies that $$A_n \xrightarrow{n \rightarrow \infty} A$$ in the operator norm. Thus, we conclude that $B(X,Y)$ is a Banach space.
As a Functional Analysis example, consider the space $X=C^0([0,1])$, the space of the continuous functions on the interval $[0,1]$. Consider the norm $\|\cdot\|_2$ on $X$ defined by $$ \|f\|_2=\left(\int_0^1|f(t)|^2\, dt\right)^{1/2}. $$ Then $(X,\|\cdot\|_2)$ is not complete. In fact, you can find a $\|\cdot\|_2$-Cauchy sequence which would converge to a discountinuous function (hence to something outside $X$). For example you can approximate (in the sense of the norm $\|\cdot\|_2$) the step function with jump at $1/2$ by menas of continuous functions. This would not be possible in the sense of the norm $\|\cdot\|_\infty$! After all, $(X,\|\cdot\|_\infty)$ is a complete normed space.
Best Answer
The answer to the question exactly as you asked it is yes; your space is isomorphic as a vector space, with no topology, to various Banach spaces. (See various comments for details.)
Edit: The assertion that the answer is yes has met with vigorous disbelief. Also there's a technical point that I realized after some thought I simply didn't know how to do. I've added an Appendix at the bottom of this answer explaining things in detail, in particular showing that $\Bbb R^{\Bbb N}$ and $\ell^2(\Bbb N)$ do in fact have algebraic dimension $c$. Back to the short version:
But that norm simply has nothing to do with the structure of the space as a space of sequences. There's no complete norm on the space $X$ of all real sequences that has the sort of properties you must have in mind. In particular: For $n\in\Bbb N$ define $\Lambda_n:X\to\Bbb R$ by $$\Lambda_n x=x_n.$$There is no complete norm on $X$ such that every $\Lambda_n$ is bounded. (That is, there is no complete norm such that for every $n$, $x_n$ depends continuously on $x$.) In fact there's no such norm, complete or not.
Suppose on the other hand that $||\Lambda_n||=c_n$. Define $$S_N=\sum_{n=1}^N 2^{-n}c_n^{-1}\Lambda_n.$$Then $||S_N||<1$ for every $N$, which says that for every $x$ we have $$\left|\sum_{n=1}^N2^{-n}c_n^{-1}x_n\right|\le||x||$$for every $N$. This is impossible; for any sequence $c_n$ there exists a sequence $x$ such that $$\sup_N\left|\sum_{n=1}^N2^{-n}c_n^{-1}x_n\right|=\infty.$$
Appendix: Two things: why those bases show that the answer to the question is yes, and why those bases exist. Note that here "basis" means basis in the pure linear algebra sense, often called "Hamel basis": If $B$ is a basis for $X$ then every $x\in X$ is equal to a unique linear combination of finitely many elements of $B$. $\newcommand\S{\Bbb R^{\Bbb N}}$ $\newcommand\L{\ell^2(\Bbb N)}$
Assume for now that $\S$ and $\L$ both have bases of cardinality $c$. Then there is a bijection betweenn the two bases. This gives us an isomorphism $I:\S\to\L$, defined by mapping linear combinations of the elements of the basis for $\S$ to the corresponding linear combinations of the elements of the basis for $\L$. We can thus define a norm on $\S$ by $||x||_{\S}=||Ix||_{\L}$, and it is straightforward to verify that the new norm on $\S$ is complete (given a Cauchy sequence in $\S$, by definition the corresponding sequence in $\L$ is Cauchy, hence convergent, so that by definition the original sequence in $\S$ is convergent.)
Ok, I left out some details there. Every detail is trivial - this equals that be definition since this or that is an isomorphism.
How do we know there are bases of cardinality $c$? Both spaces have cardinality $c$ so a basis can be no larger than $c$; we need $c$ independent vectors. It seems clear that we somehow should get this from the fact that there are $c$ subsets of $\Bbb N$, but I didn't see how to get the independence. I asked the Banach-space guy at the office:
Main Lemma There exists a map $S:\Bbb R\to\mathcal P(\Bbb N)$ such that $S(r)$ is infinite for every $r$, and such that given finitely many distinct reals $r_1,\dots,r_n$ there exists $N$ so that the sets $S(r_1)\cap[N,\infty),\dots,S(r_n)\cap[N,\infty)$ are pairwise disjoint.
Proof Let $\Bbb Q=\{q_1,\dots\}$. For each $r\in\Bbb R$ choose a sequence $(q_{n_j})$ of distinct rationals with $q_{n_j}\to r$, and let $S(r)=\{n_j\}$. QED.
Now it's clear that $\{\chi_{S(r)}:r\in\Bbb R\}$ is a linearly independent subset of $\S$. And if $(e_1,\dots)$ is orthonormal in $\L$ then $\left\{\sum_{j\in S(r)}\frac{e_j}{j}:r\in R\right\}$ is a linearly independent subset of $\L$.