About the closedness of $\frac d{dx}$ operator

Question

The example section of the closed linear operator in the wikipedia page https://en.wikipedia.org/wiki/Unbounded_operator#Closed_linear_operators says that

(1) "Consider the derivative operator $A =\frac d {dx}$ where $X = Y = C([a, b])$ is the Banach space of all continuous functions on an interval $[a, b]$. If one takes its domain $D(A)$ to be $C^1([a, b])$, then $A$ is a closed operator, which is not bounded."

(2) "On the other hand if $D(A) = C^\infty([a, b])$, then $A$ will no longer be closed, but it will be closable, with the closure being its extension defined on $C^1([a, b])$."

The first claim seems easy to understand. Since the limit of any sequence of continuously differentiable functions $f_n$ if converges would be a continuously differentiable function $f$ and $\frac d {dx} f$ is a continuous function $\in C[a,b]$. However the second claim seems not to be easy to understand. Can anyone give a more rigorously proof to this two statements clearly? Thanks.

Answer 1

Recall this theorem:

Let $I = [a,b]$ and $f_n : I \to \mathbb{R}$ a sequence of differentiable functions such that the sequence of derivatives $(f_n')_n$ converges uniformly to a function $g : I \to \mathbb{R}$. Also $\exists x_0 \in I$ such that the sequence $(f_n(x_0))_n$ converges. Then $(f_n)_n$ converges uniformly to a differentiable function $f : I \to \mathbb{R}$ with $f' = g$.

So if $(f_n)_n$ is a sequence of functions in $C^1[a,b]$ such that $f_n \to f$ uniformly in $C[a,b]$ and $f_n' \to g$ uniformly in $C[a,b]$ then the previous theorem gives that $f$ is differentiable and $f' = g$. Since $g$ is a uniform limit of continuous functions, it is continuous so $f' \in C[a,b]$ and therefore $f \in C^1[a,b]$.

Therefore the graph $\{(f,f') : f \in C^1[a,b]\}$ is closed so $\frac{d}{dx}$ is closed on $C^1[a,b]$.

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To show that $\frac{d}{dx}$ is not closed on $C^\infty[-1,1]$, consider $f_n \in C^\infty[-1,1]$ given by

$$f_n(x) = \frac12x\sqrt{x^2+\frac1n} + \frac1{2n}\ln\left(x + \sqrt{x^2+\frac1n}\right)$$

We have $f_n \to \frac12 x|x|$ uniformly. To see this, note that $x\mapsto x+\sqrt{x^2+\frac1n}$ is increasing and $x\mapsto\left|\ln x\right|$ increases when $x \ge 1$ and decreases when $x \le 1$.

Therefore

$$\left|\ln\left(x+\sqrt{x^2+\frac1n}\right)\right| \le \max\left\{-\ln\left(-1+\sqrt{1+\frac1n}\right), \ln\left(1+\sqrt{1+\frac1n}\right)\right\} \le \max\left\{-\ln\left(-1+\sqrt{1+\frac1n}\right), \ln\left(1+\sqrt2\right)\right\} \le -\ln\left(-1+\sqrt{1+\frac1n}\right)$$

for large enough $n$ and

$$\left|\sqrt{x^2+\frac1n} - |x|\right| = \sqrt{x^2+\frac1n} - \sqrt{x^2} = \frac{\frac1n}{\sqrt{x^2+\frac1n} + \sqrt{x^2}} \le \frac{\frac1n}{\frac1{\sqrt{n}}} = \frac1{\sqrt{n}}$$ so

\begin{align} \left|f_n(x) - \frac12x|x|\right| &= \left|\frac12x\sqrt{x^2+\frac1n} + \frac1{2n}\ln\left(x + \sqrt{x^2+\frac1n}\right) - \frac12 x|x|\right| \\ &\le \frac{|x|}2\left|\sqrt{x^2+\frac1n} - |x|\right| + \frac1{2n}\left|\ln\left(x + \sqrt{x^2+\frac1n}\right)\right|\\ &\le \frac1{\sqrt{n}} - \frac1{2n}\ln\left(-1+\sqrt{1+\frac1n}\right)\\ &\xrightarrow{n\to\infty} 0 \end{align}

uniformly in $x \in [-1,1]$.

On the other hand, $f_n'(x) = \sqrt{x^2+\frac1n}$ and the above computation also shows that $f_n' \to |\cdot|$ uniformly, and $|x| = \frac{d}{dx}\left(\frac12 x|x|\right)$.

Hence $f_n \to \frac12 x|x|$ uniformly and $f_n' \to \frac{d}{dx}\left(\frac12 x|x|\right)$ uniformly, but $\frac12 x|x|$ is not a smooth function.

We conclude that $\frac{d}{dx}$ is not closed on $C^\infty[-1,1]$.

Answer 2

Let us begin by recalling that an operator $A:X \to X$ with domain $D(A) \subseteq X$ is closed if its graph $$G(A) = \{(x, Ax) : x \in D(A)\} \subseteq X \times X$$ is closed.

For the first example, this means we want to check that if $f_n \in D(A) = C^1[a,b]$ and $f_n \to f$, $Af_n = f_n' \to g$ in $X = C[a,b]$ then we have that $f \in C^1[a,b]$ and $f' = g$. This is really a simple exercise is real analysis. Start by writing $$f_n(t) = f_n(a) + \int_a^t f_n'(s) ds$$ and then send $n \to \infty$ (using the fact that the convergence $f_n' \to g$ is uniform to pass the limit through the integral) to see that $$f(t) = f(a) + \int_a^t g(s) ds.$$ So $f \in C^1[a,b]$ and $f' = g$ as desired.

For the second example, it suffices to take some sequence of functions $f_n \in C^\infty[a,b]$ such that $f_n \to f$, $f_n' \to f'$ in $C[a,b]$ with $f \not \in C^\infty[a,b]$ (since then $(f_n, f_n')$ is a sequence in $G(A)$ that converges in $X \times X$ to an element of the complement of $G(A)$). To do this, you can use the fact that $C^\infty$ is dense in $C^1$ with its usual norm (simply pick any $f$ in $C^1 \setminus C^\infty$ and approximate it in $C^1$-norm by elements of $C^\infty$). This idea also tells you that any closed extension of this second operator must contain the first operator.

Hence, since the operator $A$ in the first example is a closed extension of the one in the second, it follows that this second operator is closable (since that just means that it has a closed extension) and that its closure is the operator of the first example.