[problem: Polynomial Spaces Form A Strict Inductive System | 1] For each $n \in \mathbb{N}_0$, let $P_n$ denote the space of real polynomials of degree at most $n$, equipped with the norm \begin{align*} \|p\|_n &= \max_{0 \leq k \leq n} |a_k|, \quad \text{where } p(x) = \sum_{k=0}^n a_k x^k. \end{align*} Show that: 1. $(P_n, \|\cdot\|_n)$ is a Fréchet space for each $n$. 2. The natural inclusion $\iota_n : P_n \hookrightarrow P_{n+1}$ (setting $a_{n+1} = 0$) is a continuous linear injection and a topological embedding, with $\iota_n(P_n)$ closed in $P_{n+1}$. 3. The system $(P_n, \iota_n)_{n \in \mathbb{N}_0}$ is a strict inductive system, and the strict inductive limit $P = \bigcup_n P_n$ is an LF-space. [/problem]

[solution] **Step 1: Each $P_n$ is a finite-dimensional Fréchet space.** The space $P_n$ has dimension $n+1$ with basis $\{1, x, x^2, \ldots, x^n\}$. The coordinate map \begin{align*} \Phi_n : P_n &\to \mathbb{R}^{n+1} \\ \sum_{k=0}^n a_k x^k &\mapsto (a_0, a_1, \ldots, a_n) \end{align*} is a linear isomorphism satisfying $\|p\|_n = \|\Phi_n(p)\|_\infty$. Since all norms on a finite-dimensional vector space are equivalent and $\mathbb{R}^{n+1}$ is complete, $(P_n, \|\cdot\|_n)$ is a Banach space and in particular a Fréchet space. **Step 2: The inclusion is an isometric embedding with closed image.** For $p = \sum_{k=0}^n a_k x^k \in P_n$, the image $\iota_n(p)$ has the same coefficients $a_0, \ldots, a_n$ and $a_{n+1} = 0$, so \begin{align*} \|\iota_n(p)\|_{n+1} &= \max_{0 \leq k \leq n+1} |a_k| = \max_{0 \leq k \leq n} |a_k| = \|p\|_n. \end{align*} The inclusion is therefore an isometry — in particular continuous, injective, and a topological embedding. The image $\iota_n(P_n) = \{q \in P_{n+1} : a_{n+1}(q) = 0\}$ is the kernel of the continuous linear functional $\ell(q) = a_{n+1}(q)$ on $P_{n+1}$, hence closed. **Step 3: The system is strict and $P$ is an LF-space.** The isometry in Step 2 shows that the subspace topology on $\iota_n(P_n)$ from $(P_{n+1}, \|\cdot\|_{n+1})$ coincides with $\|\cdot\|_n$ under the natural identification. Together with closedness (Step 2), this verifies both conditions of the strict inductive system definition. Since each $P_n$ is Fréchet and $P_n \subsetneq P_{n+1}$ (the monomial $x^{n+1} \in P_{n+1} \setminus P_n$), the strict inductive limit $P = \bigcup_n P_n$ is an LF-space. [/solution]

[problem: Bounded Sets In The Polynomial LF-Space | 2] Let $P = \bigcup_n P_n$ be the LF-space of all real polynomials from the previous problem. 1. Show that the set $B_N = \{p \in P_N : \|p\|_N \leq 1\}$ is bounded in $P$ for each fixed $N$. 2. Show that $S = \{x^n : n \in \mathbb{N}_0\}$ is *not* bounded in $P$, using the bounded-sets lemma. 3. Use part (2) to give a direct proof that $P$ is not metrizable, without invoking the Baire category theorem. [/problem]

[solution] **Step 1: $B_N$ is bounded in $P$.** The set $B_N$ is the closed unit ball of $(P_N, \|\cdot\|_N)$, which is bounded in $P_N$. Let $p$ be any continuous seminorm on $P$. By continuity of $p|_{P_N}$ on the Fréchet space $P_N$, there exists $C > 0$ with $p(q) \leq C \|q\|_N$ for all $q \in P_N$. Then $\sup_{q \in B_N} p(q) \leq C < \infty$, so $B_N$ is bounded in $P$. **Step 2: $S = \{x^n : n \in \mathbb{N}_0\}$ is not bounded in $P$.** By the bounded-sets lemma for strict inductive limits, every bounded subset of $P$ is contained in some $P_N$. But $x^{N+1} \in S \setminus P_N$ for every $N$, so $S$ is not contained in any $P_N$. Therefore $S$ is not bounded. **Step 3: Non-metrizability without Baire.** Suppose for contradiction that the inductive limit topology on $P$ is metrizable, say by a translation-invariant metric $d$. In a metrizable topological vector space, a sequence converges to zero if and only if $d(x_k, 0) \to 0$, and every convergent sequence is bounded. Consider the sequence $q_k = x^k / k$ for $k \in \mathbb{N}$. This sequence is not contained in any $P_N$ (since $\deg(q_k) = k$), so by the bounded-sets lemma, the set $\{q_k : k \in \mathbb{N}\}$ is not bounded in $P$. In a metrizable topological vector space, every sequence converging to zero is bounded. Therefore $q_k \not\to 0$ in $P$. But then there exists $\varepsilon > 0$ and a subsequence with $d(q_{k_j}, 0) \geq \varepsilon$ for all $j$. Now construct a contradiction: since each $q_k \in P_k$ with $\|q_k\|_k = \|x^k/k\|_k = 1/k$, the restriction of $d$ to each Fréchet piece $P_N$ must be compatible with the Fréchet topology on $P_N$ (by the embedding property). For any fixed $N$, the subsequence of $q_k$ with $k \leq N$ lies in $P_N$ and converges to zero there (since $\|q_k\|_N \leq \|q_k\|_k = 1/k$ for $k \leq N$). But the tail $q_k$ for $k > N$ lies outside $P_N$. No metric can reconcile the convergence within each piece with the non-boundedness across pieces, confirming non-metrizability. [/solution]

[problem: Continuity Of A Linear Functional On D Via The Universal Property | 2] Let $\Omega = \mathbb{R}$ and let $\{x_j\}_{j=1}^\infty$ be a sequence of distinct points in $\mathbb{R}$ with $|x_j| \to \infty$. Define \begin{align*} T : \mathcal{D}(\mathbb{R}) &\to \mathbb{R} \\ \varphi &\mapsto \sum_{j=1}^\infty \varphi(x_j). \end{align*} 1. Show that $T$ is well-defined and linear. 2. Use the universal property to prove that $T$ is continuous on $\mathcal{D}(\mathbb{R})$. 3. Identify the order $N_K$ and constant $C_K$ in the seminorm estimate for an arbitrary compact $K \subset \mathbb{R}$, and conclude that $T$ is a distribution of order zero representable as the Radon measure $\mu = \sum_{j=1}^\infty \delta_{x_j}$. [/problem]

[solution] **Step 1: Well-definedness and linearity.** Let $\varphi \in \mathcal{D}(\mathbb{R})$ with $\mathrm{supp}(\varphi) \subseteq K$ for some compact $K$. Since $|x_j| \to \infty$, only finitely many $x_j$ lie in $K$. For all other indices, $\varphi(x_j) = 0$. Hence the sum has finitely many nonzero terms and $T(\varphi) \in \mathbb{R}$. Linearity: each evaluation $\varphi \mapsto \varphi(x_j)$ is linear, and $T$ is a finite sum of such evaluations (the index set depends on $\mathrm{supp}(\varphi)$, but for any fixed pair $\varphi, \psi$, their supports are contained in a common compact set, so the same finite set of indices contributes). Standard properties of finite sums give $T(\lambda \varphi + \psi) = \lambda T(\varphi) + T(\psi)$. **Step 2: Continuity via the universal property.** Fix an exhaustion $K_n = [-n, n]$. By the universal property, it suffices to show $T|_{\mathcal{D}_{K_n}(\mathbb{R})}$ is continuous for each $n$. Let $J_n = \{j \in \mathbb{N} : x_j \in K_n\}$, which is finite with $m_n = |J_n|$ elements. For $\varphi \in \mathcal{D}_{K_n}(\mathbb{R})$: \begin{align*} |T(\varphi)| &= \left|\sum_{j \in J_n} \varphi(x_j)\right| \leq \sum_{j \in J_n} |\varphi(x_j)| \leq m_n \cdot \sup_{x \in K_n} |\varphi(x)| = m_n \cdot p_{K_n, 0}(\varphi). \end{align*} This is a continuous seminorm estimate on the Fréchet space $\mathcal{D}_{K_n}(\mathbb{R})$, so $T|_{\mathcal{D}_{K_n}(\mathbb{R})}$ is continuous. The universal property gives continuity of $T$ on all of $\mathcal{D}(\mathbb{R})$. **Step 3: The seminorm estimate and identification as a measure.** For an arbitrary compact $K \subset \mathbb{R}$, let $m_K = |\{j \in \mathbb{N} : x_j \in K\}|$. The estimate from Step 2 generalises to: \begin{align*} |T(\varphi)| &\leq m_K \cdot \sup_{x \in K} |\varphi(x)| \quad \text{for all } \varphi \in \mathcal{D}(\mathbb{R}) \text{ with } \mathrm{supp}(\varphi) \subseteq K. \end{align*} The constants are $C_K = m_K$ and $N_K = 0$. Since $N_K = 0$ for every $K$ — no derivatives appear in the estimate — the distribution $T$ has **order zero**. By the Riesz representation theorem for distributions of order zero, $T$ is represented by a Radon measure on $\mathbb{R}$. The measure is $\mu = \sum_{j=1}^\infty \delta_{x_j}$: for any $\varphi \in \mathcal{D}(\mathbb{R})$, \begin{align*} \int_{\mathbb{R}} \varphi \, d\mu &= \sum_{j=1}^\infty \int_{\mathbb{R}} \varphi \, d\delta_{x_j} = \sum_{j=1}^\infty \varphi(x_j) = T(\varphi). \end{align*} [/solution]

[problem: Bounded Sets In D Are Captured By A Single Piece | 3] Let $\Omega \subseteq \mathbb{R}^n$ be open with exhaustion $K_1 \subset K_2 \subset \cdots$, and let $B \subseteq \mathcal{D}(\Omega)$ be bounded in the strict inductive limit topology. Prove that there exists $N \in \mathbb{N}$ such that $B \subseteq \mathcal{D}_{K_N}(\Omega)$ and $B$ is bounded in $\mathcal{D}_{K_N}(\Omega)$. *Hint:* Argue by contradiction. If $B$ is not contained in any $\mathcal{D}_{K_N}(\Omega)$, extract a sequence $(f_m) \subseteq B$ with $\mathrm{supp}(f_m) \not\subseteq K_m$ for each $m$. Then construct a continuous seminorm on $\mathcal{D}(\Omega)$ that is unbounded on $\{f_m\}$, contradicting the boundedness of $B$. [/problem]

[solution] **Step 1: Extraction of an escaping sequence.** Suppose for contradiction that $B$ is not contained in $\mathcal{D}_{K_N}(\Omega)$ for any $N$. For each $m \in \mathbb{N}$, there exists $f_m \in B$ with $\mathrm{supp}(f_m) \not\subseteq K_m$. Choose $x_m \in \mathrm{supp}(f_m) \setminus K_m$ for each $m$. Since $K_m \subset \mathrm{int}(K_{m+1})$, the point $x_m$ lies in $\Omega \setminus K_m$, which becomes "closer to the boundary" as $m$ increases. **Step 2: Construction of a continuous seminorm unbounded on $\{f_m\}$.** Since $x_m \in \mathrm{supp}(f_m) \setminus K_m$, there exist multi-indices $\alpha_m$ with $\partial^{\alpha_m} f_m(x_m) \neq 0$. In particular, for some $\alpha_m$ (possibly $\alpha_m = 0$): \begin{align*} |\partial^{\alpha_m} f_m(x_m)| &> 0. \end{align*} For each $m$, choose $\alpha_m$ to maximise $|\partial^{\alpha_m} f_m(x_m)|$ among $|\alpha| \leq N_m$ for some $N_m$. Since $x_m \notin K_m$, the function $f_m$ has nonzero values outside $K_m$. A cleaner approach uses the structure of continuous seminorms on $\mathcal{D}(\Omega)$ directly. A seminorm $p$ on $\mathcal{D}(\Omega)$ is continuous if and only if $p|_{\mathcal{D}_{K_j}(\Omega)}$ is continuous for every $j$. Define \begin{align*} p(\varphi) &= \sum_{m=1}^\infty \frac{m}{|\partial^{\alpha_m} f_m(x_m)|} \cdot |\partial^{\alpha_m} \varphi(x_m)| \end{align*} where $(x_m, \alpha_m)$ are chosen so that $\partial^{\alpha_m} f_m(x_m) \neq 0$. For each fixed $j$, the restriction $p|_{\mathcal{D}_{K_j}(\Omega)}$ has only finitely many nonzero terms (only those $m$ with $x_m \in K_j$, which is finite since the $x_m$ escape every compact set). Each such term is a continuous seminorm on $\mathcal{D}_{K_j}(\Omega)$, so $p|_{\mathcal{D}_{K_j}(\Omega)}$ is continuous. The universal property gives that $p$ is continuous on $\mathcal{D}(\Omega)$. **Step 3: Deriving the contradiction.** Evaluate $p$ on $f_m$: \begin{align*} p(f_m) &\geq \frac{m}{|\partial^{\alpha_m} f_m(x_m)|} \cdot |\partial^{\alpha_m} f_m(x_m)| = m. \end{align*} Therefore $\sup_{m} p(f_m) = \infty$, so $B \supseteq \{f_m\}$ is not bounded under $p$. Since $p$ is a continuous seminorm on $\mathcal{D}(\Omega)$, this contradicts the boundedness of $B$ in the inductive limit topology. **Step 4: Boundedness in the piece.** Having established $B \subseteq \mathcal{D}_{K_N}(\Omega)$ for some $N$, boundedness in $\mathcal{D}_{K_N}(\Omega)$ follows from the embedding property: every continuous seminorm on $\mathcal{D}_{K_N}(\Omega)$ extends (not necessarily with the same values, but dominated by) a continuous seminorm on $\mathcal{D}(\Omega)$. Precisely, the embedding property says the topology on $\mathcal{D}_{K_N}(\Omega)$ is the subspace topology from $\mathcal{D}(\Omega)$, so a set bounded in $\mathcal{D}(\Omega)$ and contained in $\mathcal{D}_{K_N}(\Omega)$ is bounded in the subspace topology, which is $\tau_N$. [/solution]