As we known, Banach spaces are normed vector spaces where Cauchy sequences converge. Can someone give me some examples of vector spaces, with a defined norm, which are not complete?
[Math] Example of a non complete normed vector space.
functional-analysisnormed-spacesvector-spaces
Related Solutions
Part 1: (There are examples in which is complete under $||\cdot||_2$ and not under $||\cdot||_1$)
Take $V:=C[0,1]$, $||f||_1:=\int_{0}^{1}f^2$, and $||f||_2:=\sup|f|$. Then $||\cdot||_1\leq||\cdot||_2$. While $V$ is complete under $||\cdot||_2$ we can get a sequence of continuous functions converging to $\chi_{[1/2,1]}$ under $||\cdot||_1$.
Part 2: (It is not possible for $V$ to be complete under $||\cdot||_1$)
Assume that $V$ were complete under $||\cdot||_1$. Consider the identity function $I:(V,||\cdot||_2)\rightarrow (V,||\cdot||_1)$. Since $||I(f)||_1=||f||_1\leq K||f||_2$, we have that $I$ is a continuous linear map. Since it is also surjective (and we are assuming $(V,||\cdot||_2)$ and $(V,||\cdot||_1)$ to be Banach) we get, by the open mapping theorem, that $I$ is open.This means that $I^{-1}:(V,||\cdot||_1)\rightarrow (V,||\cdot||_2)$ is continuous. Since $I^{-1}$ is linear, then is is also bounded: $$||f||_2=||I^{-1}(f)||_2\leq C||f||_1$$ for some constant $C>0$.
Therefore $\frac{1}{C}||\cdot||_2\leq||\cdot||_1\leq K||\cdot||_1$. This means that the norms would be equivalent. But this is a contradiction with the fact that they are not. Therefore $V$ cannot be complete under $||\cdot||_1$.
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.
Best Answer
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.