If $f_k \to f$ uniformly, then $J(f_k) \to J(f)$, where $J: \mathcal{C}_{\text{c}}(\mathbb{R}^d) \to \mathbb{R}$ ist linear and monotonous

Question

Let $\mathcal{C}_{\text{c}}(\mathbb{R}^d) := \{f: \mathbb{R}^d \xrightarrow{\mathcal{C}} \mathbb{R}: \text{supp}(f) \text{ is compact.} \}$.

I have a few questions regarding the proof following

Lemma Let $J: \mathcal{C}_{\text{c}}(\mathbb{R}^d) \to \mathbb{R}$ be a linear and monotonous functional and further $f, f_1, f_2, \ldots \in \mathcal{C}_{\text{c}}(\mathbb{R}^d)$ with supp$(f_1)$, supp$(f_2), \ldots \subset Q$, where $Q \subset \mathbb{R}^d$ is a axially parallel cuboid. If $f_k$ converges uniformly to $f$, then $\lim\limits_{k \to \infty} J(f_k) = J(f)$.

Proof

Let $\Phi: \mathbb{R}^d \to [0,1]$ be in $\mathcal{C}_{\text{c}}(\mathbb{R}^d)$ with $\Phi|_Q \equiv 1$. Because the support of $f_k - f$ is in $Q$ also, we obtain \begin{equation} \tag{1} -\|f_k - f\|_{\infty} \Phi \le f_k - f \le \|f_k - f\|_{\infty} \Phi \qquad \qquad \text{in } \mathbb{R}^d \end{equation} and by linearity and monotonicity of J \begin{equation} \tag{2} -\|f_k - f\|_{\infty} J(\Phi) \le J(f_k) - J(f) \le \|f_k - f\|_{\infty} J(\Phi). \end{equation} therefore it follows that $|J(f_k) - J(f)| \le \|f_k - f\|_{\infty} J(\Phi)$ and therefore the lemma.

My Questions

1. Why is supp$(f_k - f)$ in Q, when we know nothing about supp$(f)$?

2. Where does the fist inequality come from? When we add an argument $x \in \mathbb{R}^n$ to the functions, which are not inside norms, it seems clearer, but what do we know about $\Phi(x)$ if $x \not\in Q$?

3. Why do we choose $\Phi$ like this? What is the underlying idea?

4. Does the lemma basically say that if $f_k$ converges uniformly, we can exchange a linear monotonous function with limit?

Before you answer Please note, this is the first proof from my analysis 3 course, meaning I am familiar with differentiation in multiple dimensions and one dimensional integration as well as linear algebra but nothing more.

Answer 1

1. You have $f_k(x) = 0$ for $x \notin Q$, hence $f(x) = \lim_k f_k(x) = 0$ for $x \notin Q$.

2. You only need to know that $\Phi(x) = 1$ for $x \in Q$. Then for $x \in Q$ $$(1) \phantom{xx} -\|f_k - f\|_{\infty} \Phi(x) = -\|f_k - f\|_{\infty} \le f_k(x) - f(x) \le \|f_k - f\|_{\infty}$$ $$ = \|f_k - f\|_{\infty} \Phi(x)$$ and for $x \notin Q$ $$(2) \phantom{xx} -\|f_k - f\|_{\infty} \Phi(x) \le 0 = f_k(x) - f(x) \le \|f_k - f\|_{\infty} \Phi(x) .$$

3. The idea is that we need to apply $J$ to elements of $\mathcal{C}_c(\mathbb{R}^d)$. $f_k$ and $f$ as well as $f_k - f$ are such elements, and the constant $\|f_k - f\|_{\infty}$ is transformed into such an element by multiplication with $\Phi$. Here the essential point is that $\Phi(x) = 1$ for $x \in Q$, otherwise you cannot be sure that (1) is satisfied.

4. It says that $J$ is continuous: $\lim_k f_k = f$ in the metric space $(\mathcal{C}_c(\mathbb{R}^d),\lVert - \rVert_\infty)$ implies $\lim_k J(f_k) = J(f)$ in the metric space $(\mathbb{R},\lvert-\rvert)$. Or, if you want, $J(\lim_k f_k) = \lim_k J(f_k)$.

Answer 2

1. You know that for any $k\in\mathbb{N}$ it holds: $$\forall x\in \mathbb{R}^d\backslash Q: f_k(x)= 0$$ So given an arbitrary $\varepsilon>0$ uniform convergence implies $$\forall x\in \mathbb{R}^d\backslash Q: |f_k(x)-f(x)|= |f(x)|<\varepsilon$$ Therefore you must have: $$\forall x\in\mathbb{R}^d\backslash Q: f_k(x)-f(x )=f(x)=0 $$

2. The first inequality is actually the same as the second. $$f-f_k, f_k - f \le \|f_k - f\|_{\infty} \Phi $$ implies by multiplication with $-1$ $$ -\|f_k - f\|_{\infty} \Phi \le f_k-f,f-f_k $$ at any point in $\mathbb{R}^d$.