- [Convergence, Completeness in Metric Space](Functional%20Spaces/Convergence,%20Completeness%20in%20Metric%20Space.md) - [Hahn Banach Theorem and its Corollaries](Hahn%20Banach%20Theorem%20and%20its%20Corollaries.md) - [Hilbert Space Introduction](Hilbert%20Space%20Introduction.md) --- ### **Intro** Let $X$ be Hilbert space. So it's a special kind of Metric space. Recall that a sequence $x_n$ converges to $x$ in the metric means that $d(x, x_n) \rightarrow 0$ where $d(x, x_n)$ is just a real number. This is also referred to as strong convergence in Hilbert space. In Hilbert space, strong convergence of $x \rightarrow x$ is equivalent to $\lim_{x\rightarrow \bar x}\Vert x - \bar x\Vert = 0$. However, this is not the only mode of convergences in Hilbert space. We define weak convergence next. #### **Def | Weakly Converging Sequence** > Let $x_n$ be a sequence in $X$ that converges weakly to a point $x$, denote as $x_n \rightharpoonup x$, then for all $v \in X$ we have $\langle x_n, v\rangle\rightarrow 0$. **Observations** A sequence the converges weakly is not a sequence that converges strongly. Obviously consider the standard basis vector of $l^2$ with $\e_i$. We have $\langle \e_i, v\rangle = v_i$ for all $v_i$, for any $v \in X$. Since $v \in X$, we must have the partial sum $\sum_{i = 1}^{n} v_i^2$ to be convergent. And therefore $v_i$ is bounded by $\epsilon^2_i$, therefore it converges weakly. Also it doesn't converge strongly because it's not Cauchy. We have $\Vert \e_i - \e_j\Vert = 1$ for all $i \neq j$. **Remarks** In finite dimension, weak convergence is the same as strong convergence. #### **Def | Weak Seqeuentially Closed** > A set $D \subseteq X$ is a sequential weakly closed set if, for all $x_n \in X$ such that $x_n \rightharpoonup x$, then $x \in D$. > #### **Facts** > We state a list of useful facts that may be useful. > 1. Weak limits if exists they are unique. > 2. Strong convergence implies weak convergence. > 3. $x_n \rightarrow x \iff (x_n \rightharpoonup x \wedge \Vert x _n \Vert\rightarrow \Vert x\Vert)$ > 4. $x_n \rightharpoonup x$ implies that $x_n$ is a bounded vector. > 5. Weak convergence is able to achieve convergence point-wise. Meaning that if $x_n \rightharpoonup y$, then $x_{n, k}\rightarrow y_{n, k}$ for all $k \in \mathbb N$. > 6. Every bounded sequence has a weakly convergent subsequence. **Remarks** (6.) is the Bozano Weierstrass theorem manifested in the infinite dimensional Hilbert Space. --- ### **Theorems** We prove some theorems about weak convergence. They are useful for applying the concepts in various context. #### **Thm | The Weak Limit is Unique** > Let $x_n \rightharpoonup y$ and $x_n\rightharpoonup y$, then $y = z$. **Proof** Consider any $p \in X$, then consider the difference as part of the definition of the weak convergence of the sequence: $ \begin{aligned} \langle x_n - z, p\rangle - \langle x_n - y, p\rangle &= \langle y - z, p\rangle \\ \lim_{n\rightarrow \infty} \left( \langle x_n - z, p\rangle - \langle x_n - y, p\rangle \right) &= \langle y - z, p\rangle \\ \lim_{n\rightarrow \infty}\langle x_n - z, p\rangle - \lim_{n\rightarrow \infty}\langle x_n - y, p\rangle &= \langle y - z, p\rangle \\ 0 &= \langle y - z, p\rangle, \end{aligned} $ in which case, we have that $\langle y - z, p\rangle$ for all $p \in X$. Therefore $y = z$. We can break the limit apart in the above derivation because weak convergence asserts the existence of the limits of both parts. #### **Thm | Equivalent Conditions for Strong Convergence** > Let $x_n \in X$ be a sequence in Hilbert space then$x_n \rightarrow x$ if and onlyif $x \rightharpoonup x$ and $\Vert x_n \Vert\rightarrow \Vert x\Vert$. **Proof** Direction ($\implies$) is direct. Because the mapping $\Vert x_n\Vert$ is a continuous mapping, and by property of metric space, the limit would be $\Vert x\Vert$. For proving $(\impliedby)$, consider $ \begin{aligned} \Vert x_n - x\Vert^2 &= \Vert x_n\Vert - \Vert x\Vert^2 - 2\langle x, x_n\rangle \\ &= \Vert x_n\Vert^2 + \Vert x\Vert^2 - 2(\langle x, x_n - x\rangle + \Vert x\Vert^2) \\ \implies \lim_{n\rightarrow \infty} \Vert x_n - x\Vert^2 &= \lim_{n\rightarrow \infty} \Vert x_n\Vert^2 + \Vert x \Vert^2 - 2\left(\lim_{n\rightarrow \infty} \langle x, x_n - x\rangle + \Vert x\Vert^2\right) \\ \text{by } & \lim_{n\rightarrow \infty} \Vert x_n\Vert^2 = \Vert x\Vert^2 \\ \implies &= \Vert x\Vert^2 + \Vert x\Vert^2 - 2(0 + \Vert x\Vert^2) \\ &= 0. \end{aligned} $ therefore, the sequence converges strongly. #### **Thm | Convergence of the Dot Product** > Let $x_n\rightarrow x$ and $y_n \rightharpoonup y$, then $\langle x_n, y_n\rangle\rightarrow \langle x, y\rangle$. **Observations** The dot product between the dot product of 2 weakly converging sequence is not the same as the dot product of the limit of individual weak sequence. Observe the counter example that $x_i = y_n = \e_i \rightharpoonup \mathbf 0$, but their dot product would be $\Vert \e_1\Vert = 1$, which is not the dot product of their limits $\langle \mathbf 0, \mathbf 0\rangle = 0$. **Proof** To prove, for this proof we will assume that a weakly converging sequence is bounded. This claim is not trivial to show. Recall from definition that $x_n\rightarrow x \iff \lim_{n\rightarrow \infty}\langle x_n - x, x_n - x\rangle = 0$. And for the weak limit we have $y_n\rightharpoonup y \iff \lim_{n\rightarrow \infty}\langle y_n - y, z\rangle = 0$. With that we consider $ \begin{aligned} \langle x_n, y_n\rangle &= \langle x_n, y_n - y\rangle + \langle x_n, y\rangle \\ &= \langle x_n - x, y_n - y\rangle + \langle x, y_n - y\rangle + \langle x_n, y\rangle \\ &\le \Vert x_n - x\Vert \Vert y_n - y\Vert + \langle x, y_n - y\rangle + \langle x_n, y\rangle \end{aligned} $ takingthe limit, we have the term $\Vert x_n - x\Vert\Vert y_n - y\Vert \rightarrow 0$ because $\Vert x_n - x\Vert\rightarrow 0$ and $\Vert y_n - y\Vert < \infty$. The cross term $\langle x, y_n - y\rangle$ is also zero by the weak convergence of $y_n$. Therefore, the above expression as limit $\langle x_n, y\rangle\rightarrow \langle x, y\rangle$ by the strong convergence of sequence $x_n$.