Banach Spaces

Banach Spaces and Convergence of Series

A norm tells us how to measure the size of vectors and the distance between them; completeness tells us whether Cauchy sequences have limits in the space. The previous tutorial established examples of complete and incomplete normed spaces. This tutorial develops a useful way to recognize completeness: in a Banach space, a series whose terms have summable norms converges to an element of the space.

Definition: A Banach space is a normed vector space $(V,N)$ that is complete: every Cauchy sequence in $V$, measured using $N$, converges in norm to an element of $V$. A series $\sum_{n=1}^{\infty}x_n$ in $V$ is called absolutely convergent if $\sum_{n=1}^{\infty}N(x_n)$ is a convergent series of real numbers.

The word “absolutely” refers to the sum of the norms, not to an order or sign on the vectors. In $\mathbb{R}$, it corresponds to the familiar condition $\sum |x_n|<\infty$. In a general normed space, summability of the sizes $N(x_n)$ controls the tails of the vector series through the triangle inequality.

A Series Characterization of Completeness

Completeness gives a guarantee about every Cauchy sequence. The following theorem gives an equivalent guarantee in terms of series. Its reverse direction is especially useful: carefully selected successive differences of a Cauchy sequence form an absolutely convergent series.

Theorem (Series Characterization of Banach Spaces): A normed space $V$ is complete if and only if every series $\sum_{n=1}^{\infty}x_n$ in $V$ satisfying $\sum_{n=1}^{\infty}N(x_n)<\infty$ converges in $V$.

Proof. First suppose $V$ is complete, and let $\sum_{n=1}^{\infty}x_n$ satisfy $\sum_{n=1}^{\infty}N(x_n)<\infty$. Define the partial sums $s_m=\sum_{n=1}^{m}x_n$. If $m>n$, the triangle inequality gives $$ N(s_m-s_n) =N\left(\sum_{j=n+1}^{m}x_j\right) \leq\sum_{j=n+1}^{m}N(x_j) \leq\sum_{j=n+1}^{\infty}N(x_j). $$ The last expression tends to zero as $n$ tends to infinity. Thus $(s_m)$ is Cauchy in $V$. Completeness supplies an $s\in V$ such that $s_m\to s$ in norm. By definition, the series converges to $s$.

Conversely, suppose every series with summable norms converges in $V$, and let $(y_n)$ be any Cauchy sequence in $V$. We construct a subsequence with successive differences small enough to have summable norms. For each positive integer $k$, the Cauchy property gives an index $M_k$ such that $$ m,n\geq M_k\quad\Longrightarrow\quad N(y_m-y_n)<2^{-k}. $$ Choose indices recursively so that $n_1\geq M_1$, and, for $k\geq2$, $$ n_k\geq\max\{M_k,n_{k-1}+1\}. $$ The indices are strictly increasing, and $n_k\geq M_k$ for every $k$. In particular, both $n_k$ and $n_{k+1}$ are at least $M_k$. Therefore $$ N(y_{n_{k+1}}-y_{n_k})<2^{-k}. $$ It follows that $$ \sum_{k=1}^{\infty}N(y_{n_{k+1}}-y_{n_k}) \leq\sum_{k=1}^{\infty}2^{-k}<\infty. $$ By the assumed series property, the series of differences converges in $V$. Thus its partial sums converge, and $$ y_{n_1}+\sum_{k=1}^{r}(y_{n_{k+1}}-y_{n_k})=y_{n_{r+1}} $$ shows that the subsequence $(y_{n_k})$ converges to some $y\in V$.

It remains to show that the full Cauchy sequence converges to the same $y$. Let $\varepsilon>0$. Choose $N$ so that $m,n\geq N$ implies $N(y_m-y_n)<\varepsilon/2$. Since $n_k$ tends to infinity and $y_{n_k}\to y$, choose $k$ with $n_k\geq N$ and $N(y_{n_k}-y)<\varepsilon/2$. For every $n\geq N$, the triangle inequality yields $$ N(y_n-y)\leq N(y_n-y_{n_k})+N(y_{n_k}-y)<\varepsilon. $$ Hence $y_n\to y$ in $V$. Every Cauchy sequence in $V$ converges in $V$, so $V$ is complete. $\square$

The subsequence selection is essential. A Cauchy threshold $M_k$ controls a pair of terms only when both indices are at least $M_k$. Requiring $n_k\geq M_k$ at every stage ensures that the two indices in the $k$-th successive difference meet that threshold. Once the subsequence converges, the Cauchy property transfers its limit to the full sequence.

Using the Series Criterion in Examples

Worked Example: A Uniformly Convergent Series of Continuous Functions

Consider $C[0,1]$ with the supremum norm, which is complete by the Completeness of $C[a,b]$ in the Supremum Norm. For $n\geq1$, define $$ u_n(x)=\left(\frac{x}{3}\right)^n,\qquad 0\leq x\leq1. $$ Each $u_n$ is continuous, and $$ \|u_n\|_\infty=\sup_{0\leq x\leq1}\left(\frac{x}{3}\right)^n=\left(\frac13\right)^n. $$ The equality holds because $0\leq x\leq1$, so $(x/3)^n\leq(1/3)^n$, with equality at $x=1$. The geometric series of norms converges: $$ \sum_{n=1}^{\infty}\|u_n\|_\infty =\sum_{n=1}^{\infty}\left(\frac13\right)^n =\frac12. $$ The Series Characterization of Banach Spaces therefore implies that $\sum_{n=1}^{\infty}u_n$ converges in the supremum norm to an element of $C[0,1]$. Its value can also be calculated pointwise. For $0\leq x\leq1$, the geometric-series formula gives $$ \sum_{n=1}^{\infty}\left(\frac{x}{3}\right)^n =\frac{x/3}{1-x/3} =\frac{x}{3-x}. $$ The denominator $3-x$ is at least $2$ on $[0,1]$, so this formula defines a continuous function there. The norm estimate establishes convergence in the space; the formula identifies the limit.

Worked Example: An Absolutely Convergent Series in the Plane

In $\mathbb{R}^2$ with its Euclidean norm, let $$ x_n=\left(\frac{(-1)^n}{3^n},\frac{2}{5^n}\right),\qquad n\geq1. $$ For real coordinates $a,b$, $\sqrt{a^2+b^2}\leq |a|+|b|$, since squaring both nonnegative sides reduces the inequality to $0\leq2|a||b|$. Thus $$ \|x_n\|_2 =\sqrt{\frac{1}{9^n}+\frac{4}{25^n}} \leq\frac{1}{3^n}+\frac{2}{5^n}. $$ Both geometric series on the right converge, so $\sum_{n=1}^{\infty}\|x_n\|_2<\infty$. Since $\mathbb{R}^2$ is complete, the vector series converges. Its coordinates can be summed separately: $$ \sum_{n=1}^{\infty}\frac{(-1)^n}{3^n}=-\frac14, \qquad \sum_{n=1}^{\infty}\frac{2}{5^n}=\frac12. $$ For the first equality, the geometric ratio is $-1/3$, so the sum is $(-1/3)/(1-(-1/3))=-1/4$. For the second, the sum is $2(1/5)/(1-1/5)=1/2$. Hence the vector sum is $(-1/4,1/2)$.

Completeness Does Not Depend on the Choice of an Equivalent Norm

A vector space can be equipped with more than one norm. Some norms may measure distances differently while still controlling one another up to fixed positive constants. Such norms give the same Cauchy sequences and the same convergent sequences, and consequently have the same completeness property.

Definition: Two norms $N_1$ and $N_2$ on the same vector space $V$ are equivalent if there are constants $c,C>0$ such that $$ cN_1(x)\leq N_2(x)\leq CN_1(x)\qquad\text{for every }x\in V. $$

Theorem (Completeness Is Invariant Under Equivalent Norms): If $N_1$ and $N_2$ are equivalent norms on $V$, then $(V,N_1)$ is complete if and only if $(V,N_2)$ is complete.

Proof. Suppose $(V,N_1)$ is complete, and let $(x_n)$ be Cauchy with respect to $N_2$. The lower comparison $cN_1(x)\leq N_2(x)$ gives $$ N_1(x_n-x_m)\leq\frac{1}{c}N_2(x_n-x_m). $$ As $m,n$ tend to infinity, the right-hand side tends to zero, so $(x_n)$ is Cauchy with respect to $N_1$. Completeness of $N_1$ gives an $x\in V$ with $N_1(x_n-x)\to0$. The upper comparison then gives $$ N_2(x_n-x)\leq C N_1(x_n-x)\longrightarrow0. $$ Thus $(V,N_2)$ is complete. Interchanging the roles of the norms, or using $N_1(x)\leq N_2(x)/c$ and $N_2(x)\leq C N_1(x)$, gives the reverse implication. $\square$

Worked Example: An Equivalent Norm on Continuous Functions

On $C[0,1]$, define $$ N_*(f)=\|f\|_\infty+|f(0)|. $$ This is a norm: nonnegativity and homogeneity follow from those of the absolute value and supremum norm; if $N_*(f)=0$, then $\|f\|_\infty=0$, hence $f=0$; and the triangle inequality follows by adding the two triangle inequalities for its terms. Moreover, for every $f\in C[0,1]$, $$ \|f\|_\infty\leq N_*(f)\leq2\|f\|_\infty, $$ because $|f(0)|\leq\|f\|_\infty$. Thus $N_*$ and the supremum norm are equivalent. Since $C[0,1]$ is complete in the supremum norm, the Invariance Theorem shows it is complete in $N_*$ as well. The value at the endpoint changes the measured size, but only by a controlled amount; it does not change whether Cauchy sequences have limits.

An Incomplete Space Detected by a Series

The series criterion also gives a direct way to see how completeness can fail. Let $c_{00}$ be the vector space of real sequences with only finitely many nonzero coordinates, equipped with the norm $$ \|a\|_1=\sum_{j=1}^{\infty}|a_j|. $$ This sum is finite for every $a\in c_{00}$. The following sequence of finite-support vectors is Cauchy in this norm, but its limit does not have finite support.

Worked Example: Finite-Support Sequences Need Not Form a Banach Space

For each positive integer $N$, define $v^{(N)}\in c_{00}$ by $$ v^{(N)}_j= \begin{cases} 3^{-j},&1\leq j\leq N,\\ 0,&j>N. \end{cases} $$ If $M>N$, then $$ \|v^{(M)}-v^{(N)}\|_1 =\sum_{j=N+1}^{M}3^{-j} \leq\sum_{j=N+1}^{\infty}3^{-j} =\frac{3^{-N}}{2}. $$ The last expression tends to zero with $N$, so $(v^{(N)})$ is Cauchy. Suppose it converged in $c_{00}$ to some $v$. For each fixed coordinate $j$, the coordinate difference is bounded by the norm: $$ |v_j-v^{(N)}_j|\leq\|v-v^{(N)}\|_1. $$ For $N\geq j$, $v^{(N)}_j=3^{-j}$. Taking $N$ to infinity would therefore give $v_j=3^{-j}$ for every $j$. This is impossible for $v\in c_{00}$, since $3^{-j}\neq0$ for every positive integer $j$, so $v$ would have infinitely many nonzero coordinates. Thus the Cauchy sequence has no limit in $c_{00}$, and this normed space is not complete.

The same failure is visible in the series of coordinate vectors: its finite partial sums are exactly the $v^{(N)}$, and the norms of its successive terms are $3^{-j}$, whose sum is finite. The Series Characterization says that this behavior cannot occur in a complete space. Here the series has no sum in $c_{00}$, even though its partial sums are Cauchy.

What the Banach Space Viewpoint Adds

A Banach space is more than a space in which limits happen to exist. Its completeness lets estimates on approximations produce an actual element of the space. The series criterion is one particularly efficient form of this principle: summable bounds on successive changes ensure that the partial sums are Cauchy, and completeness supplies their limit. Conversely, the proof shows that the ability to sum every series with summable norms is strong enough to recover convergence of every Cauchy sequence.

Equivalent norms provide a second useful perspective. Completeness is not sensitive to changes in the norm that are bounded above and below by fixed positive multiples. But an arbitrary change of norm need not have this property; the space and its norm must be considered together. The example of $c_{00}$ illustrates why: its Cauchy approximations approach an object that the space excludes.

Takeaway: A normed space is a Banach space precisely when every series with summable norms converges in that space. Equivalent norms preserve completeness, while a Cauchy sequence or an absolutely convergent series whose limit is missing can reveal incompleteness.

Check Your Understanding

Use the definitions and proofs in this tutorial to answer the following questions.

What does it mean for a series in a normed space to be absolutely convergent?
Why do the partial sums of a series with summable norms form a Cauchy sequence?
In the reverse direction of the Series Characterization, why must each selected index $n_k$ be at least $M_k$?
What inequalities show that $N_*(f)=\|f\|_\infty+|f(0)|$ is equivalent to the supremum norm?
Why can the Cauchy sequence $v^{(N)}$ not converge to an element of $c_{00}$?

What You'll Learn