1. Show that
\[ \langle x, y \rangle = \frac14 (||x + y||^2 - ||x - y||^2) \]
Solution:
\[ \begin{aligned} ||x + y||^2 &= ||x||^2 + ||y||^2 + 2\langle x, y \rangle \\ ||x - y||^2 &= ||x||^2 + ||y||^2 - 2\langle x, y \rangle \\ ||x + y||^2 - ||x - y||^2 &= 4\langle x, y \rangle \\ \langle x, y \rangle &= \frac14 (||x + y||^2 - ||x - y||^2) \end{aligned} \]
2. Show that (parallelogram identity)
\[ ||x + y||^2 + ||x - y||^2 = 2(||x||^2 + ||y||^2) \]
Solution: Adding together \(||x + y||^2\) and \(||x - y||^2\) from 1. we see that
\[ \begin{aligned} ||x + y||^2 +||x - y||^2 &= 2(||x||^2 + ||y||^2) \end{aligned} \]
3. Show that \(f: H\times H \rightarrow \mathbb R, f(x,y) = \langle x, y \rangle\) is a continuous mapping, i.e. show that \(x_n, y_n \rightarrow x, y\) in H implies \(\langle x_n, y_n \rangle \rightarrow \langle x, y \rangle\) in \(\mathbb R\)
Solution:
\[ \begin{aligned} \langle x_n, y_n \rangle &= \langle x_n + x - x, y_n + y - y \rangle \\ &= \langle x_n - x, y_n + y - y \rangle + \langle x, y_n + y - y \rangle \\ &= \langle x_n - x, y_n - y \rangle + \langle x_n - x, y \rangle + \langle x, y_n + y - y \rangle \\ &= \langle x_n - x, y_n - y \rangle + \langle x_n - x, y \rangle + \langle x, y_n- y \rangle + \langle x, y \rangle \\ &\rightarrow_{n\rightarrow \infty} \langle 0, 0 \rangle + \langle 0, y \rangle + \langle x, 0 \rangle + \langle x, y \rangle \\ &= \langle x, y \rangle \end{aligned} \]
4. Show that \(H \rightarrow \mathbb R, x \rightarrow ||x||\) is a continuous mapping.
Solution: Setting \(y = x\) and \(y_n = x_n\) in the previous solution proves this statement.
5. Show that if H is a Hilbert space over \(\mathbb C\), then
\[ \langle x, y \rangle = \frac 14 (||x + y||^2 - ||x - y||^2 + i||x + iy||^2 - i||x - iy||^2) \]
Solution:
\[ \begin{aligned} ||x + y||^2 &= ||x||^2 + ||y||^2 + \langle x, y \rangle + \overline{\langle x, y \rangle} \\ ||x - y||^2 &= ||x||^2 + ||y||^2 - \langle x, y \rangle - \overline{\langle x, y \rangle} \\ i||x + iy||^2 &= i\langle x + iy, x + iy \rangle \\ &= i\langle x, x + iy \rangle + i\langle iy, x + iy \rangle \\ &= i\overline{\langle x + iy, x \rangle} - \overline{\langle x + iy, y \rangle} \\ &= i\overline{\langle x, x \rangle} + i\overline{\langle iy, x \rangle} - \overline{\langle x, y \rangle} - \overline{\langle iy, y \rangle} \\ &= i||x||^2 + i\bar i \overline{\langle y, x \rangle} - \overline{\langle x, y \rangle} - \bar i ||y||^2 \\ &= i||x||^2 i ||y||^2 + \langle x, y \rangle - \overline{\langle x, y \rangle} \\ i||x - iy||^2 &= i\langle x - iy, x - iy \rangle \\ &= i\langle x, x - iy \rangle + \langle y, x - iy \rangle \\ &= i\overline{\langle x - iy, x \rangle} + \overline{\langle x - iy, y \rangle} \\ &= i\overline{\langle x, x \rangle} - i\overline{\langle iy, x \rangle} + \overline{\langle x, y \rangle} - \overline{\langle iy, y \rangle} \\ &= i||x||^2 - \langle x, y \rangle + \overline{\langle x, y \rangle} + i||y||^2 \end{aligned} \]
Adding all the above terms together we get the desired solution.
6. Which of the statements below hold true if H is merely a pre-Hilbert space.
A subset \(X \subset H\) is said to be convex if \(x,y\in X\) and \(t\in[0, 1]\) implies \(tx + (1-t)y \in X\).
A subset \(X\) is said to be closed if \(\{x_n\}_{n\in \mathbb N} \subset X, x_n \rightarrow x\) implies \(x\in X\). Not complete.
For a linear subspace \(X \subset H\) we write \(x\perp X\) iff \(\langle x, y \rangle = 0 \ \forall y \in X.\) We define \(X^{\perp} := \{x\in H : x\perp X\}\).
1. Show that a linear subspace \(X \subset H\) is convex.
Solution: Set \(u = \lambda \in R, w = 1 - \lambda\) and \(x, y\in X\). Since \(x, y \in X\) then \(ux, wy \in X\). Moreover, since \(ux, wy \in X\) we have that \(ux + wy \in X\) or \(\lambda x + (1 - \lambda) y \in X\) which proves the statement.
2. Show that if \(\lambda \in H^*\) then \(ker(\lambda) = \{x\in H : \lambda(x) = 0\}\) is closed.
Solution: Since \(\lambda \in H^*\) we know that \(\lambda\) is linear and continuous. We know that \(\{0\}\) is closed and thus \(ker(\lambda)\) is also closed.
3. Show that if \(X \subset H\) is a linear subspace, then \(X^\perp\) is closed.
Solution: Let \(y, z \in X^\perp, a,b\in \mathbb F, x \in X\). Then \(\langle ay + bz, x \rangle = a\langle y, x \rangle + b\langle z, x \rangle = 0\) so \(ay + bz \in X^\perp\) and \(X^\perp\) thus a linear subspace of H. Now let \(\{w_n\}\) be a sequence in \(X^\perp\) and \(w_n \rightarrow w \in X\) when \(n \rightarrow \infty\). Then
\[ \lim_{n\rightarrow\infty}\langle w - w_n, x \rangle = \langle w, x \rangle - \lim_{n\rightarrow\infty}\langle w_n, x \rangle = \langle w, x \rangle = 0 \] since \(w_n \in X^\perp, x \in X\). Thus \(w \in X^\perp\) and \(X^\perp\) is closed.
4. Assume that \(X \subset H\) is a closed linear subspace of H. Show that for every \(y \in H\) there exists only one \(x \in X\) such that \(\langle y - x, z \rangle = 0\) for all \(z \in X\). Further, show that for this x we have \(||y - z|| = inf_{s\in X}||y - s||\).
Solution: If \(y \in X\) we simply choose y = x and there is nothing left to show. Let \(d = inf_{x\in X}||y - x||\), \(d > 0\) since \(y \not \in X\) and X is closed. Consider the sequence \(\{x_n\}_{n=1}^\infty\) in X such that \(||y - x_n|| \rightarrow d\) as \(n \rightarrow \infty\) and \(d^2 \leq ||y - x_n|| < d^2 + \frac 1n\) We need to show that this is a Cauchy sequence. Using the parallelogram law with \(A = y - x_n, B = y - x_m\) and the fact that for each \(n \in \mathbb N\) we can find an \(x_n \in X\).
\[ \begin{aligned} ||A + B||^2 +||A - B||^2 &= 2(||A||^2 + ||B||^2) \\ \rightarrow ||(y - x_n) + (y - x_m)||^2 + ||(y - x_n) - (y - x_m)||^2 &= 2(||y - x_n||^2 + ||y - x_m||^2) \\ \rightarrow ||2y - (x_n + x_m)||^2 + ||x_n - x_m||^2 &= 2(||y - x_n||^2 + ||y - x_m||^2) \\ &\leq 4d^2 + 2(\frac 1n + \frac 1m) \end{aligned} \]
Since X is a linear subspace with \(x_n, x_m \in X\) we see that \(\frac12 (x_n + x_m) \in X\) so
\[ ||2y - (x_n + x_m)||^2 = 4||y - \frac12 (x_n + x_m)||^2 \geq 4d^2 \]
which gives
\[ \begin{aligned} ||x_n - x_m||^2 &= 2(||y - x_n||^2 + ||y - x_m||^2) - ||2y - (x_n + x_m)||^2 \\ &\leq 4d^2 + 2(\frac 1n + \frac 1m) - 4d^2 \\ &= 2(\frac 1n + \frac 1m) \end{aligned} \]
Hence \(\{x_n\}\) is Cauchy and coverges to some \(x \in X\) since X is closed and \(d^2 \leq \lim_{n\rightarrow \infty}||y - x_n|| = || y - x|| \leq d^2\) and so the required \(x\) exists. To prove uniqueness, soppose that \(z \in X, ||y - z|| = d\). Since X is a linear subspace \(\frac12 (q + w) \in X\) and \(||y - \frac12(x + z)|| \geq d\). Applying the parallelogram rule to \(y - x\) and \(y - z\) we get
\[ \begin{aligned} ||(y - x) + (y - z)||^2 + ||(y - x) - (y - z)||^2 &= 2||y - x||^2 + 2||y - z||^2 \\ \rightarrow ||x - z||^2 &= 2d^2 + 2d^2 - 4||y - \frac12(x + z)||^2 \\ &\leq 2d^2 + 2d^2 - 4d^2 = 0 \end{aligned} \]
so x = z.
1. Assume that \(X \subset H\) is a closed linear subspace of H. Show that every \(x \in H\) can be written uniquely as \(x = x^\parallel + x^\perp\), where \(x^\parallel \in X\) and \(x^\perp \in X^\perp\).
Solution: Set \(x \in H\). Problem 2.3 shows us that there is a \(y \in X\) such that for all \(u \in X\), \(||x - y|| \leq ||x - u||\). Let \(z = x - y\). Then for all \(u \in X\),
\[ \begin{aligned} ||z - u|| = ||x - y - u|| = ||x - (y + u)|| \geq ||x - y|| = ||z|| \\ ||z|| - ||z - u|| \geq 0 \end{aligned} \]
So \(z \in X^\perp\). Since \(y \in X, z \in X^\perp\) we see that \(x = x^\parallel + x^\perp\) which proves existance. To prove uniqueness, suppose \(x = y_1 + z_1 = y_2 + z_2, y_1, y_2 \in X, z_1, z_2 \in X^\perp\). Then \(y_1 - y_2 = z_2 - z_1\) but \(y_1 - y_2 \in X\) and \(z_2 - z_2 \in X^\perp\), so \(y_1 - y_2 = z_2 - z_1 \in X \cap X^\perp = \{0\}\) so \(y_1 = y_2, z_1 = z_2\).
2. Assume that \(X \subset H\) is a closed linear subspace of H. Show that \(||x||^2 = ||x^\parallel||^2 + ||x^\perp||^2\).
Solution:
\[ ||x||^2 = ||x^\parallel + x^\perp||^2 = \langle x^\parallel + x^\perp, x^\parallel + x^\perp \rangle = ||x^\parallel||^2 + ||x^\perp||^2 + \langle x^\parallel, x^\perp \rangle + \langle x^\perp, x^\parallel \rangle = ||x^\parallel||^2 + ||x^\perp|| \]
3. Prove the Riesz representation theorem without the assumption of separability.
Solution: If f(x) = 0 for all \(x \in H\) then y = 0 because \(\langle x, 0 \rangle = 0\). Otherwise, \(ker(f) = \{x\in H: f(x) = 0\}\) is a closed subspace of H which means that \(ker(f)^\perp \neq \{0\}\). There fore there exists \(z \in ker(f)^\perp\) such that \(f(z) = 1\). Since \(z\neq 0\) we define \(y = \frac{z}{||z||^2}\). Let \(x\in H\). Since \(f\) is linear we have that
\[ f(x - f(x)z) = f(x) - f(x)f(z) = f(x) - f(x) = 0 \] and thus \(x - f(x)z \in ker(f)\). We also have that \(z \in ker(f)^\perp\) so
\[ \begin{aligned} \langle x - f(x)z, z \rangle &= \langle x, z \rangle - f(x)\langle z, z \rangle = 0\\ &\rightarrow \langle x, z \rangle = f(x)||z||^2 \\ &\rightarrow f(x) = \frac{1}{||z||^2}\langle x, z \rangle = \langle x, \frac{z}{||z||^2} \rangle \\ &= \langle x, y \rangle \end{aligned} \]
If \(y\) and \(w\) are such that \(f(x) = \langle x, y \rangle = \langle x, w \rangle\) for all \(x \in H\), then \(\langle x, y - w \rangle = 0\) for all \(x \in H\). This means that \(y - w = 0\) and \(y = w\).
Now let H be a Hilbert space of functions \(\Omega \rightarrow \mathbb R\).
1. Show that for every \(x \in \Omega\) the mapping \(L_x: H \rightarrow \mathbb R, L_x(f) = f(x)\) is linear.
Solution: Set \(a, b \in R, x, y, z \in \Omega, f(x) = \langle x, z \rangle\)
\[ \begin{aligned} L_{ax + by}(f) &= f(ax + by) \end{aligned} \]
2. Show that if for every \(x \in \Omega\) there exists a \(M_x \geq 0\) such that \(|L_x(f)| \leq M_x||f|\) for all \(f \in H\), then \(L_x \in H^*\)
Solution: Since we know that \(L\) is linear we only need to show that is is continuous to prove it is part of \(H^*\). Any linear function that is continuous at 0 is continuous everywhere so we prove that \(L\) is continuous at 0. Assume a sequence \(\{f_n\}\) where \(f_n \rightarrow 0\) when \(n \rightarrow \infty\). Then \(||L_x(f_n)|| \leq M_x||f_n|| \rightarrow 0\), which means that \(L_x(f_n) \rightarrow 0\) so \(L\) is continuous at zero.
3. Show that for every \(x \in \Omega\) there exists a \(K_x \in H\) such that \(f(x) = \langle f, K_x \rangle\) for all \(f\in H\).
Solution:
4. Show that \(K: \Omega \times \Omega \rightarrow \mathbb R, K(x, y) = \langle K_x, K_y \rangle\), defines a function.
Solution:
5. Show that for \(x_1, x_2, ..., x_n \in \Omega\) and \(c_1, c_2, ..., c_n \in \mathbb R\) we have \(\sum_{i,j=1}^nc_ic_jK(x_i,x_j)\geq0.\)
Solution: