Function-Space Proof Workshop

Why a Function Space Needs the Right Norm

A norm on a function space determines what it means for functions to be close. In the supremum norm, closeness controls function values uniformly, but it says nothing directly about derivatives. If a problem requires both the functions and their derivatives to behave well, the norm should measure both. This tutorial develops a standard example: the space of continuously differentiable functions with a norm that controls the function and its derivative together.

Throughout, let \(a<b\). Write \(C[a,b]\) for the continuous real-valued functions on \([a,b]\), with the supremum norm \(\|f\|_\infty=\sup_{x\in[a,b]}|f(x)|\). The completeness of \(C[a,b]\) in this norm was established earlier in “Complete Normed Spaces.” We will use that result to prove a new completeness theorem for a function space with stronger control.

The two terms have distinct roles. The first controls the size of \(f\); the second controls the size of its rate of change. The sum is convenient because it controls each term separately: \(\|f\|_\infty\leq\|f\|_{C^1}\) and \(\|f'\|_\infty\leq\|f\|_{C^1}\). In particular, convergence in the \(C^1\) norm implies uniform convergence of both the functions and their derivatives.

Checking the Norm and Using It

Proof. Each term is nonnegative, so \(\|f\|_{C^1}\geq0\). If \(\|f\|_{C^1}=0\), then \(\|f\|_\infty=0\), which implies \(f(x)=0\) for every \(x\in[a,b]\). Conversely, the zero function has \(C^1\) norm zero. For a scalar \(c\), we have \((cf)'=cf'\), and hence \(\|cf\|_{C^1}=|c|\|f\|_\infty+|c|\|f'\|_\infty=|c|\|f\|_{C^1}\).

For \(f,g\in C^1([a,b])\), the triangle inequality for the supremum norm gives \(\|f+g\|_\infty\leq\|f\|_\infty+\|g\|_\infty\). Since \((f+g)'=f'+g'\), it also gives \(\|(f+g)'\|_\infty\leq\|f'\|_\infty+\|g'\|_\infty\). Adding these inequalities yields \(\|f+g\|_{C^1}\leq\|f\|_{C^1}+\|g\|_{C^1}\). All norm axioms hold. \(\square\)

The derivative itself is a linear map from \(C^1([a,b])\) into \(C[a,b]\). The choice of norm makes this map bounded, a useful structural feature rather than a coincidence.

Proof. Differentiation is linear, so \(D(\alpha f+\beta g)=\alpha D(f)+\beta D(g)\) for scalars \(\alpha,\beta\). Also, \(\|D(f)\|_\infty=\|f'\|_\infty\leq\|f\|_\infty+\|f'\|_\infty=\|f\|_{C^1}\). This is the boundedness inequality with constant \(1\). \(\square\)

Completeness of the \(C^1\) Space

A normed space is complete if every Cauchy sequence converges to an element of that same space. Completeness matters in proofs because it guarantees that a sequence whose terms become arbitrarily close has a limit that remains an admissible object. For \(C^1([a,b])\), the key point is to track the derivatives as well as the functions.

Proof. Let \((f_n)\) be Cauchy in the \(C^1\) norm. For any \(\varepsilon>0\), there is an \(N\) such that, for \(m,n\geq N\),

Each term on the left is nonnegative, so each is less than \(\varepsilon\). Therefore \((f_n)\) is Cauchy in \(C[a,b]\), and \((f_n')\) is also Cauchy in \(C[a,b]\). By completeness of \(C[a,b]\), there exist \(f,g\in C[a,b]\) such that \(f_n\to f\) uniformly and \(f_n'\to g\) uniformly.

We must show that \(f\) has derivative \(g\); uniform convergence of two sequences alone does not establish that relationship. For each \(n\) and each \(x\in[a,b]\), the Fundamental Theorem of Calculus gives

We can pass to the limit in this identity. First, uniform convergence implies \(f_n(x)\to f(x)\) and \(f_n(a)\to f(a)\). For the integral, for every \(x\in[a,b]\),

Taking limits in the identity therefore gives \(f(x)=f(a)+\int_a^x g(t)\,dt\) for every \(x\in[a,b]\). Since \(g\) is continuous, the Fundamental Theorem of Calculus implies that \(f\) is differentiable on \((a,b)\) and \(f'=g\) there. The derivative extends continuously to \([a,b]\) as \(g\), so \(f\in C^1([a,b])\).

Finally, \[ \|f_n-f\|_{C^1}=\|f_n-f\|_\infty+\|f_n'-f'\|_\infty. \] Both terms tend to zero: the first because \(f_n\to f\) uniformly, and the second because \(f_n'\to g=f'\) uniformly. Hence \(f_n\to f\) in the \(C^1\) norm. Every Cauchy sequence in this space converges within it, proving completeness. \(\square\)

Why the Supremum Norm Alone Is Not Enough

The completeness theorem depends on controlling derivatives. If \(C^1([a,b])\) is instead given only the supremum norm, uniform limits may no longer be differentiable. The next example makes the distinction precise on \([-1,1]\).

There is no contradiction with the completeness of \(C^1([a,b])\) in its \(C^1\) norm. The sequence in the example is Cauchy only with respect to function values; it does not satisfy the additional Cauchy control on derivatives required by the \(C^1\) norm. A norm is not just a label for a space: changing it can change which sequences converge and whether the space is complete.

The third step is the essential safeguard in many function-space arguments. Convergence of \(f_n\) and convergence of \(f_n'\) do not by themselves prove that the two limits are a function and its derivative. The integral identity supplies the missing link. The same proof pattern appears whenever a space is defined by multiple pieces of data: establish convergence of each piece, then prove that the limiting pieces still satisfy the defining relationship.