[problem] **Problem 1.** Let $U = (0, 1)$. Consider the sequence \begin{align*} u_n: U &\to \mathbb{R} \\ x &\mapsto n\, \mathbb{1}_{(0,\, 1/n)}(x). \end{align*} 1. Does $u_n$ converge pointwise $\mathcal{L}^1$-almost everywhere? 2. Does $u_n$ converge strongly in $L^1(U)$? 3. Does $u_n$ converge weakly in $L^1(U)$? [/problem]

[solution] **Step 1: Pointwise convergence.** For any fixed $x \in (0,1)$, choose $N \in \mathbb{N}$ with $1/N < x$. For all $n > N$, the point $x$ lies outside $(0, 1/n)$, so $u_n(x) = 0$. Thus $u_n(x) \to 0$ for every $x \in (0,1)$, and the sequence converges to zero $\mathcal{L}^1$-almost everywhere. **Step 2: Strong convergence.** We compute $\|u_n\|_{L^1(U)} = \int_0^1 n\, \mathbb{1}_{(0,1/n)} \, d\mathcal{L}^1 = n \cdot \frac{1}{n} = 1$. If $u_n \to u$ strongly, the limit must agree with the pointwise limit almost everywhere, forcing $u = 0$. But $\|u_n - 0\|_{L^1} = 1 \not\to 0$, so strong convergence fails. **Step 3: Weak convergence.** The dual of $L^1(U)$ is $L^\infty(U)$. Testing against the constant function $g \equiv 1 \in L^\infty(U)$, \begin{align*} \int_0^1 u_n \cdot 1 \, d\mathcal{L}^1 = 1 \quad \text{for all } n. \end{align*} Were $u_n \rightharpoonup 0$ weakly, this integral should converge to $0$. Since $1 \neq 0$, weak convergence to zero fails. Moreover, zero is the only possible weak limit (any other candidate $u \in L^1$ would have to equal the pointwise limit $0$ almost everywhere, by testing against indicators of intervals and using a density argument). Hence $u_n$ does not converge weakly in $L^1(U)$. The mass concentrates at the origin rather than dispersing — a signature of the non-reflexivity of $L^1$. The [Weak Sequential Compactness](/theorems/214) theorem does not apply because $L^1$ is not reflexive. By contrast, the [Banach-Alaoglu theorem](/theorems/212) guarantees that $u_n$, viewed as a bounded sequence in $(C_0(U))^*$ via the Riesz-Markov theorem, has a weak* convergent subsequence. The limit is the Dirac mass $\delta_0$, which is a measure but not an $L^1$ function — the mass has escaped $L^1$ entirely. [/solution]

[problem] **Problem 2.** Let $H$ be a Hilbert space and let $\{x_n\}_{n=1}^\infty \subseteq H$ satisfy $x_n \rightharpoonup x$ weakly and $\|x_n\|_H \to \|x\|_H$. Show that $x_n \to x$ strongly. [/problem]

[solution] **Step 1: Expand the norm of the difference.** Using the inner product structure of $H$, expand \begin{align*} \|x_n - x\|_H^2 &= \langle x_n - x, x_n - x \rangle = \|x_n\|_H^2 - 2\operatorname{Re}\langle x_n, x \rangle + \|x\|_H^2. \end{align*} **Step 2: Evaluate the inner product term.** Since $x_n \rightharpoonup x$ weakly, the functional $y \mapsto \langle y, x \rangle$ (which belongs to $H^*$ by the Riesz Representation Theorem) applied to $x_n$ gives $\langle x_n, x \rangle \to \langle x, x \rangle = \|x\|_H^2$. **Step 3: Combine.** Substituting the hypothesis $\|x_n\|_H \to \|x\|_H$ and the result of Step 2: \begin{align*} \|x_n - x\|_H^2 &= \|x_n\|_H^2 - 2\langle x_n, x \rangle + \|x\|_H^2 \to \|x\|_H^2 - 2\|x\|_H^2 + \|x\|_H^2 = 0. \end{align*} Hence $x_n \to x$ strongly in $H$. This is the Hilbert space case of the Kadec-Klee property. The argument uses the inner product structure in an essential way: in a general Banach space, weak convergence plus norm convergence does not imply strong convergence without additional geometric conditions such as uniform convexity. [/solution]