# How is it that the derivative operator is a closed linear operator?

By definition, if the derivative operator $D:C^1[-1,1]\to C^1[-1,1]$ is closed, then it should be the case that, given any sequence $\{x_n\}$ in $C^1[-1,1]$, and given that $x_n\to x$ as $n\to\infty$ for some $x\in C[-1,1]$, we should conclude that $x$ is in $C^1[-1,1]$. But it seems that this is not the case.

Consider $x_n(t)=\sqrt{n^{-2}+x^2}$, clearly $x_n$ are in $C^1[-1,1]$, and we know that $x_n\to x(t)=|x|$, which is not in $C^1[-1,1]$, and doesn't it show that $D$ is not a closed linear operator?

• Convergence in which sense? – Davide Giraudo Dec 17 '12 at 19:11
• @DavideGiraudo: Under the usual norm, for example, $\|\bullet\|_\infty$? – xzhu Dec 17 '12 at 19:12

As a map from $\mathcal{C}^1$ to itself, the derivative operator is not even defined, since not every element of $\mathcal{C}^1$ is twice continuously differentiable. The derivative operator is closed from $\mathcal{C}^1$ to $\mathcal{C}^0$, with respect to the standard norms $\|f\|_{\mathcal{C}^1} = \sup |f| + \sup |f'|$ and $\|f\|_{\mathcal{C}^0} = \sup |f|$.

EDIT: The derivative operator from $\mathcal{C}^1$ to $\mathcal{C}^0$ is actually bounded by $1$, and thus also closed. However, the usual meaning of "closed operator" in functional analysis is more general, see Wikipedia. Given Banach spaces $X$ and $Y$, a possibly unbounded operator $A: \mathcal{D}(A) \to Y$ with domain $\mathcal{D}(A) \subset X$ is said to be closed if for every sequence $(x_n)$ in $\mathcal{D}(A)$, converging to $x \in X$, such that $(Ax_n)$ converges to $y$, one has that $x \in \mathcal{D}(A)$ and $Ax = y$. In that sense the derivative operator is closed on $\mathcal{C}^0$ and on $\mathcal{C}^1$, because in either case the uniform convergence of the derivatives implies that the derivative of the limit is the limit of the derivatives. Your counterexample does not work because $(x_n')$ does not converge in $\mathcal{C}^0$.

Limit of derivative is not always derivative of limit.

Consider $D: C^1[0,1] \to C^0[0,1]$ and $f_n(x)=(1/n)\sin(n \pi x)$ which is in $C^1[0,1]$.

$D[f_n(x)]=\pi \cos(n \pi x)$ which is in $C^0[0,1]$.

The function sequence $f_n$ converges to $0$ in sup norm as $n$ tends to infinity. $D[f_n]$ however does not.

• You should and must learn and use $\LaTeX$-notation. – kjetil b halvorsen Feb 5 '15 at 18:22