From Real-Valued Functions to Functions with Values in a Normed Space
The bounded real-valued functions on a nonempty set form a complete space under the supremum norm, as established earlier in this course. The same question can be asked when the values of a function are not real numbers but elements of another normed space. The answer depends exactly on whether that target space is complete.
This extension separates two sources of possible difficulty. The supremum norm measures errors uniformly over the domain, while completeness of the target guarantees that pointwise Cauchy sequences of values have limits. We will show that these two ingredients together give completeness of the function space—and that completeness of the function space cannot hold if the target itself is incomplete.
The supremum in this definition is finite precisely because \(f\) is bounded. The formula gives a norm on \(B(E;V)\): nonnegativity and homogeneity follow from the corresponding properties of \(N_V\), and \(\|f\|_\infty=0\) implies \(N_V(f(x))=0\) for every \(x\), hence \(f(x)=0\) everywhere. For the triangle inequality, for every \(x\in E\), $$ N_V(f(x)+g(x))\leq N_V(f(x))+N_V(g(x))\leq \|f\|_\infty+\|g\|_\infty. $$ Taking the supremum over \(x\) gives \(\|f+g\|_\infty\leq\|f\|_\infty+\|g\|_\infty\). Thus \(B(E;V)\) is a normed space.
Completeness of the Bounded Function Space
The following theorem identifies the precise relationship between the target and the function space. In the forward direction, a Cauchy sequence of functions has Cauchy values at each point. Completeness of \(V\) supplies those pointwise limits, and the uniform Cauchy estimates show that the resulting function is bounded and is the uniform limit. The reverse direction uses constant functions to place a copy of \(V\) inside the function space.
Proof. First suppose \(V\) is complete, and let \((f_n)\) be a Cauchy sequence in \(B(E;V)\). For each fixed \(x\in E\) and all \(m,n\), $$ N_V(f_n(x)-f_m(x))\leq\|f_n-f_m\|_\infty. $$ The right-hand side tends to zero as \(m,n\) tend to infinity. Therefore \((f_n(x))\) is Cauchy in \(V\). Since \(V\) is complete, there is an element \(f(x)\in V\) such that \(f_n(x)\to f(x)\). This defines a function \(f:E\to V\).
We must first verify that \(f\) is bounded. By the Cauchy property, choose an index \(N\) such that $$ \|f_m-f_N\|_\infty<1\qquad\text{for every }m\geq N. $$ For each fixed \(x\), this implies \(N_V(f_m(x)-f_N(x))<1\) for every \(m\geq N\). As \(m\) tends to infinity, \(f_m(x)\to f(x)\). Continuity of the norm, which follows from the reverse triangle inequality, gives $$ N_V(f(x)-f_N(x))\leq1. $$ Consequently, $$ N_V(f(x))\leq N_V(f_N(x))+1\leq\|f_N\|_\infty+1 $$ for every \(x\in E\). The bound on the right is finite and independent of \(x\), so \(f\in B(E;V)\).
It remains to establish convergence in the supremum norm. Let \(\varepsilon>0\). Since \((f_n)\) is Cauchy, choose \(N_\varepsilon\) such that $$ \|f_n-f_m\|_\infty<\varepsilon/2\qquad\text{whenever }n,m\geq N_\varepsilon. $$ Fix \(n\geq N_\varepsilon\) and \(x\in E\). For every \(m\geq N_\varepsilon\), $$ N_V(f_n(x)-f_m(x))<\varepsilon/2. $$ Letting \(m\) tend to infinity gives \(N_V(f_n(x)-f(x))\leq\varepsilon/2\). This holds for every \(x\), so $$ \|f_n-f\|_\infty\leq\varepsilon/2<\varepsilon. $$ Thus \(f_n\to f\) in \(B(E;V)\), proving that \(B(E;V)\) is complete.
Conversely, suppose \(B(E;V)\) is complete. Choose one point \(x_0\in E\), which is possible because \(E\) is nonempty. Let \((v_n)\) be a Cauchy sequence in \(V\), and define the constant functions \(c_n:E\to V\) by \(c_n(x)=v_n\). They belong to \(B(E;V)\), and $$ \|c_n-c_m\|_\infty=N_V(v_n-v_m). $$ Hence \((c_n)\) is Cauchy in \(B(E;V)\). By completeness, it converges to some \(F\in B(E;V)\). For every \(x\in E\), $$ N_V(F(x)-v_n)\leq\|F-c_n\|_\infty\longrightarrow0. $$ In particular, \(v_n\to F(x_0)\). For any other \(x\in E\), both \(F(x)\) and \(F(x_0)\) are limits in \(V\) of the sequence \((v_n)\). Uniqueness of limits in a normed space gives \(F(x)=F(x_0)\). Thus \(F\) is constant, and \(v_n\) converges in \(V\). Every Cauchy sequence in \(V\) therefore converges, so \(V\) is complete. \(\square\)
Examples of the Completeness Criterion
Worked Example: Bounded Maps into the Plane
Equip \(\mathbb{R}^2\) with the maximum norm \(N_V(a,b)=\max\{|a|,|b|\}\), and let \(E=\mathbb{N}=\{1,2,3,\ldots\}\). The space \(\mathbb{R}^2\) is complete in this norm: a Cauchy sequence of pairs has Cauchy coordinate sequences in \(\mathbb{R}\), and their limits form its limit in the maximum norm. The theorem therefore guarantees that \(B(\mathbb{N};\mathbb{R}^2)\) is complete.
For a concrete sequence, define $$ f_n(k)=\left(\frac1k+\frac1n,\frac{(-1)^k}{k+n}\right),\qquad k\in\mathbb{N}. $$ Each \(f_n\) is bounded: its first coordinate has absolute value at most \(1+1/n\leq2\), and its second has absolute value at most \(1/(n+1)\leq1/2\). Define \(f(k)=(1/k,0)\), which is also bounded. For every \(k\), $$ N_V(f_n(k)-f(k)) =\max\left\{\frac1n,\frac1{k+n}\right\} =\frac1n, $$ because \(k+n\geq n+1\), so \(1/(k+n)<1/n\). Taking the supremum over \(k\) gives \(\|f_n-f\|_\infty=1/n\), which tends to zero. This calculation exhibits uniform convergence in the function-space norm.
Worked Example: Bounded Maps into a Space of Continuous Functions
Let \(V=C[0,1]\) with the supremum norm. This target space is complete (as proved in a later tutorial); the example below only uses that its elements are bounded continuous functions. Take \(E=\mathbb{N}\), and for each \(n\geq1\) define \(F_n:\mathbb{N}\to C[0,1]\) by $$ F_n(k)(t)=\frac{t}{k+n},\qquad 0\leq t\leq1. $$ For each fixed \(n\), this is a bounded map into \(C[0,1]\), since $$ \|F_n(k)\|_\infty=\sup_{0\leq t\leq1}\frac{t}{k+n}=\frac1{k+n}\leq\frac1{n+1}. $$ Here the supremum over \(t\) is attained at \(t=1\). The zero map is the uniform limit, because $$ \|F_n-0\|_\infty =\sup_{k\in\mathbb{N}}\|F_n(k)\|_\infty =\sup_{k\in\mathbb{N}}\frac1{k+n} =\frac1{n+1}. $$ The last supremum is attained at \(k=1\). This example uses two supremum norms: one measures the size of each continuous function on \([0,1]\), and the outer one measures uniformly over the index \(k\).
What Happens When the Target Is Incomplete?
The nonempty-domain hypothesis in the theorem matters for its reverse direction: constant functions on \(E\) provide a copy of the target only when there is a point at which to evaluate them. If the target is incomplete, the theorem says that its bounded function space is incomplete as well. The next example makes the failure explicit.
Worked Example: Bounded Functions into an Incomplete Target
Let \(c_{00}\) be the vector space of real sequences with only finitely many nonzero coordinates, equipped with the supremum norm $$ N_V(a)=\sup_{j\geq1}|a_j|. $$ Take \(E=\{x_0\}\), a one-point set. For each positive integer \(N\), define \(a^{(N)}\in c_{00}\) by $$ a^{(N)}_j= \begin{cases} 1/j,&1\leq j\leq N,\\ 0,&j>N. \end{cases} $$ Regard \(a^{(N)}\) as a function \(g_N:E\to c_{00}\) by setting \(g_N(x_0)=a^{(N)}\). If \(M>N\), then the coordinates of \(a^{(M)}-a^{(N)}\) are zero except for \(N<j\leq M\), where they equal \(1/j\). Thus $$ \|g_M-g_N\|_\infty =N_V(a^{(M)}-a^{(N)}) =\max_{N<j\leq M}\frac1j =\frac1{N+1}. $$ This tends to zero as \(N\) tends to infinity, so \((g_N)\) is Cauchy in \(B(E;c_{00})\).
Suppose it converged to some \(g\in B(E;c_{00})\), and write \(a=g(x_0)\in c_{00}\). For every coordinate \(j\), $$ |a_j-a^{(N)}_j|\leq N_V(a-a^{(N)})=\|g-g_N\|_\infty. $$ For all \(N\geq j\), \(a^{(N)}_j=1/j\). Since the right-hand side tends to zero, \(a_j=1/j\) for every \(j\). But this sequence has infinitely many nonzero coordinates, contradicting \(a\in c_{00}\). Hence \((g_N)\) has no limit in \(B(E;c_{00})\), and that function space is incomplete.
Closed Targets and a Common Pitfall
The theorem also combines naturally with closed subspaces. If \(V\) is complete and \(W\) is a closed linear subspace of \(V\), then \(W\) is complete in the inherited norm by the Closed Subspaces of Complete Normed Spaces theorem. Applying the completeness criterion to \(W\) gives completeness of \(B(E;W)\). Equivalently, one can view these functions as \(V\)-valued functions whose values are required to lie in \(W\).
Proof. Let \((f_n)\) be Cauchy in \(B(E;W)\). Since \(W\) is closed in the Banach space \(V\), it is complete. The Completeness of Bounded Function Spaces theorem applied with target \(W\) gives a bounded function \(f:E\to W\) such that \(\|f_n-f\|_\infty\to0\). Therefore \(f\in B(E;W)\), and the sequence converges in that space. \(\square\)
A common mistake is to assume that uniform Cauchy behavior alone guarantees a limit in any prescribed target. It guarantees that the values \(f_n(x)\) are Cauchy for each \(x\), but those values have limits in the target only if the target is complete. The \(c_{00}\) example shows that the pointwise limits can exist in a larger completion while falling outside the original target. Conversely, a complete target does not require the domain \(E\) to have any special structure: the proof used only that \(E\) is nonempty.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- Why does a Cauchy sequence in \(B(E;V)\) produce a Cauchy sequence of values at each fixed point \(x\in E\)?
- In the proof of completeness, why is a separate argument needed to show that the pointwise limit function is bounded?
- How do constant functions show that completeness of \(B(E;V)\) implies completeness of \(V\)?
- For the functions \(F_n\) into \(C[0,1]\), which two suprema are used to compute \(\|F_n\|_\infty\)?
- Why does the Cauchy sequence in \(B(\{x_0\};c_{00})\) have no limit in that space?