Compactness of a parametrized set in an infinite dimensional space.

Question

I have a question regarding the compactness of a certain parametrized set in a infinite dimensional Hilbert space.

1. Setup:

Let $\mathcal{H}$ be a real seperable Hilbert space with scalar product $\langle \cdot, \cdot \rangle$ and induced norm $\lVert \cdot \rVert$. Further, let $\mathcal{Q} \subset \mathbb{R}^m$ be an arbitrary compact set.

Consider a mapping $$ x: (0, \overline{\varepsilon}] \times \mathcal{Q} \to \mathcal{H}, \quad (\varepsilon, q) \mapsto x(\varepsilon, q), $$ with the following properties:

1. $x$ is continuous on $(0, \overline{\varepsilon}] \times \mathcal{Q}$.
2. $x(\varepsilon,q)$ is unifomrly bounded, i.e., $\exists M \ge 0$ s.t. $\lVert x(\varepsilon, q) \rVert < M $ for all $(\varepsilon, q) \in (0, \overline{\varepsilon}] \times \mathcal{Q}$.
3. For all $q \in \mathcal{Q}$ the limit $\lim_{\varepsilon \to 0} x(\varepsilon, q) =: x^*(q) \in \mathcal{H}$ exists.

In this context I have the following questions.

2. Question:

a) I would like to know if the set $$ X := \left\lbrace x(\varepsilon, q) : (\varepsilon, q) \in (0, \overline{\varepsilon}] \times \mathcal{Q}\right\rbrace $$ is relatively compact, i.e., if the closure of the set $X$ is compact in $\mathcal{H}$.

b) A weaker (sequential) version of the upper statement would also be enough. I would like to know if the following statement is true:

For every sequence $\left\lbrace \varepsilon_k \right\rbrace_{k \ge 0} \subset (0, \overline{\varepsilon}]$ and every sequence $\left\lbrace q^k \right\rbrace_{k \ge 0} \subset \mathcal{Q}$ the sequence $\left\lbrace x(\varepsilon_{k}, q^{k}) \right\rbrace_{k \ge 0}$ has a cluster point, i.e., there exists a (strictly monotonically increasing) sequence $\left\lbrace k_l \right\rbrace_{l \ge 0} \subset \mathbb{N}$ such that $\lim_{l \to + \infty} x(\varepsilon_{k_l}, q^{k_l}) = x^{\infty} \in \mathcal{H}$ exists.

I am thankful for any answer, counter example, hint to the literature, or just your general gut feeling on this.

My feeling says the statement holds and I want it to hold but so far I have not been able to show it.

Kind regards

AverageJoe

Answer 1

The set is not necessarily relatively compact, as the following example shows:

Take $\mathcal{H} = l^2(\mathbb{R})$ and $\{e_n\}_{n \geq 1}$ be the standard basis. For each $\epsilon \in (0, 1]$, let $e(\epsilon) \in \mathcal{H}$ be defined by,

$$e(\epsilon) = \cos(\frac{\pi}{2} \cdot \frac{\frac{1}{n} - \epsilon}{\frac{1}{n} - \frac{1}{n+1}}) e_n + \sin(\frac{\pi}{2} \cdot \frac{\frac{1}{n} - \epsilon}{\frac{1}{n} - \frac{1}{n+1}}) e_{n+1}$$

when $\frac{1}{n+1} < \epsilon \leq \frac{1}{n}$. That is, $e(\epsilon)$ rotates from $e_n$ to $e_{n+1}$ as $\epsilon$ decreases from $\frac{1}{n}$ to $\frac{1}{n+1}$.

Now, let $\mathcal{Q} = [0, 1]$. Define, for $\epsilon \in (0, 1]$, $q \in [0, 1]$,

$$x(\epsilon, q) = \begin{cases} e(\epsilon)&, \text{ if }\epsilon > q\\ \frac{\epsilon}{q}e(\epsilon)&, \text{ if }\epsilon \leq q \end{cases}$$

Then $x$ is continuous on $(0, 1] \times [0, 1]$, $\|x(\epsilon, q)\| \leq 1$ for all $\epsilon$ and $q$, and for any fixed $q$, $\lim_{\epsilon \to 0} x(\epsilon, q) = 0$. However, $X = \text{range}(x)$ is not relatively compact, since $e_n = x(\frac{1}{n}, \frac{1}{2n}) \in X$ for all $n \geq 1$. This also shows that the sequential version you want is also false, though that is really the same thing as sequential compactness and compactness are equivalent in a metric space.

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As requested, the following is a counterexample showing that an additional assumption that $\langle x^\ast(q) - x(\epsilon, q), x(\epsilon, q)\rangle \geq 0$ for all $\epsilon$ and $q$ is not enough to guarantee relative compactness either.

Again, let $\mathcal{H} = l^2$. This time we number the standard basis starting at 0, i.e., the standard basis shall be given by $\{e_n\}_{n \geq 0}$. We define $e(\epsilon)$ as before,

$$e(\epsilon) = \cos(\frac{\pi}{2} \cdot \frac{\frac{1}{n} - \epsilon}{\frac{1}{n} - \frac{1}{n+1}}) e_n + \sin(\frac{\pi}{2} \cdot \frac{\frac{1}{n} - \epsilon}{\frac{1}{n} - \frac{1}{n+1}}) e_{n+1}$$

when $\frac{1}{n+1} < \epsilon \leq \frac{1}{n}$, $n \geq 1$. We observe that because we start numbering the standard basis at 0, we have $e(\epsilon) \perp e_0$ for all $\epsilon$. We also note that $e(\epsilon)$ is a unit vector for all $\epsilon$.

Now, we define,

$$x(\epsilon, q) = \begin{cases} \frac{1}{2}e_0&, \text{ if }\epsilon > q\\ \frac{1}{2}e_0 + \frac{1}{2} \cdot \frac{q - \epsilon}{q/2}e(\epsilon)&, \text{ if }\frac{q}{2} < \epsilon \leq q\\ (1 - \frac{\epsilon}{q})e_0 + \sqrt{(1 - \frac{\epsilon}{q})\frac{\epsilon}{q}}e(\epsilon)&, \text{ if }\epsilon \leq \frac{q}{2} \end{cases}$$

I’ll leave it to you to check that this is indeed continuous on $(0, 1] \times [0, 1]$. Furthermore, you can calculate that,

$$x^\ast(q) = \lim_{\epsilon \to 0} x(\epsilon, q) = \begin{cases} \frac{1}{2}e_0&, \text{ if }q = 0\\ e_0&, \text{ otherwise} \end{cases}$$

When $q = 0$, $\langle x^\ast(q) - x(\epsilon, q), x(\epsilon, q)\rangle = 0$ for all $\epsilon$. If $q > 0$, we verify the three cases in the definition of $x(\epsilon, q)$ one-by-one: if $\epsilon > q$, then,

$$\langle x^\ast(q) - x(\epsilon, q), x(\epsilon, q)\rangle = \langle \frac{1}{2}e_0, \frac{1}{2}e_0\rangle = \frac{1}{4}$$

If $\frac{q}{2} < \epsilon \leq q$,

$$\langle x^\ast(q) - x(\epsilon, q), x(\epsilon, q)\rangle = \langle \frac{1}{2}e_0 - \frac{1}{2} \cdot \frac{q - \epsilon}{q/2}e(\epsilon), \frac{1}{2}e_0 + \frac{1}{2} \cdot \frac{q - \epsilon}{q/2}e(\epsilon)\rangle$$

Since $e(\epsilon)$ is a unit vector orthogonal to $e_0$ and $0 \leq \frac{q - \epsilon}{q/2} < 1$, we have,

$$\langle \frac{1}{2}e_0 - \frac{1}{2} \cdot \frac{q - \epsilon}{q/2}e(\epsilon), \frac{1}{2}e_0 + \frac{1}{2} \cdot \frac{q - \epsilon}{q/2}e(\epsilon)\rangle = \frac{1}{4} - \frac{1}{4} \cdot (\frac{q - \epsilon}{q/2})^2 \geq 0$$

That is, $\langle x^\ast(q) - x(\epsilon, q), x(\epsilon, q)\rangle \geq 0$. Finally, if $\epsilon \leq \frac{q}{2}$, then, again as $e(\epsilon)$ is a unit vector orthogonal to $e_0$,

$$\langle x^\ast(q) - x(\epsilon, q), x(\epsilon, q)\rangle = \langle \frac{\epsilon}{q}e_0 - \sqrt{(1 - \frac{\epsilon}{q})\frac{\epsilon}{q}}e(\epsilon), (1 - \frac{\epsilon}{q})e_0 + \sqrt{(1 - \frac{\epsilon}{q})\frac{\epsilon}{q}}e(\epsilon)\rangle = (1 - \frac{\epsilon}{q})\frac{\epsilon}{q} - (1 - \frac{\epsilon}{q})\frac{\epsilon}{q} = 0$$

This verifies that the example satisfies the additional assumption. However, $X = \text{range}(x)$ is still not relatively compact. Indeed, for any $n \geq 1$, $X \ni x(\frac{1}{2n}, \frac{1}{n}) = \frac{1}{2}e_0 + \frac{1}{2}e(\frac{1}{2n}) = \frac{1}{2}e_0 + \frac{1}{2}e_{2n}$.