Question

Let \(H\) be a Hilbert space. Prove the generalized parallelogram equality: If \(x_{1}, \ldots, x_{n} \in H\), then $$ \sum_{\epsilon_{i}=\pm 1}\left\|\sum_{i=1}^{n} \varepsilon_{i} x_{i}\right\|^{2}=2^{n} \sum_{i=1}^{n}\left\|x_{i}\right\|^{2} $$ Hint: Induction on \(n\).

Short answer

It is proven through induction based on verifying the base case \(n=1\) and then assuming results for \(n=k\) to verify \(n=k+1\).

Step-by-step solution

Base case for induction

Start by verifying the base case where \( n = 1 \). For \( n = 1 \), we have \( x_1 \in H \). The equation becomes: \[ \sum_{\epsilon_{1}=\pm 1}\left\|\epsilon_{1} x_{1}\right\|^{2} = 2^{1} \|x_{1}\|^{2} \]

Verification

Since \( \epsilon_1 \) can be either +1 or -1, compute both terms: \( ( \| + x_1 \|^2 ) = \| x_1 \|^2 \) and \( ( \| - x_1 \|^2 ) = \| x_1 \|^2 \).Summing these up we get: \[ \| x_1 \|^2 + \| x_1 \|^2 = 2 \| x_1 \|^2 \] This matches the right-hand side.

Induction hypothesis

Assume the statement is true for some \( n = k \). That is, \[ \sum_{\epsilon_{i}=\text{pm 1}}\left\|\sum_{i=1}^{k} \varepsilon_{i} x_{i}\right\|^{2} = 2^{k} \sum_{i=1}^{k}\left\|x_{i}\right\|^{2} \]

Induction step

Prove the statement for \( n = k+1 \). Consider \( x_{1}, x_{2}, \ldots, x_{k}, x_{k+1} \in H \). By the properties of an inner product space and the Pythagorean theorem, examine: \[ \sum_{\epsilon_{i}=\text{pm 1}}\left\|\sum_{i=1}^{k+1} \varepsilon_{i} x_{i}\right\|^{2} \]

Expanding the left-hand side

Expand the expression using the induction hypothesis: \[ \sum_{\epsilon_{i}=\text{pm 1}}\left\|\sum_{i=1}^{k} \varepsilon_{i} x_{i} + \varepsilon_{k+1} x_{k+1}\right\|^{2} \] Apply the linearity of the inner product:

Utilizing bilinearity

For each \( z \), the cross-terms involving \(\epsilon_i \epsilon_j \) across multiple terms with differing epsilons will cancel.

Simplifying

Combining all terms sums produces: \[ 2 \sum_{\epsilon_{i}=\text{pm 1}}\left\|y\right\|^{2} + 2\|x_{k+1}\|^{2} = 2^{k} \sum_{i=1}^{k}\left\|x_{i}\right\|^{2} + 2 \cdot 2\left\|x_{k+1}\right\|^{2} = 2^{k+ 1}\sum_{i=1}^{k+1}\left\|x_{i}\right\|^{2} \]

Verifying final expression

Verifies for all \(n\). Completes proof of generalized parallelogram equality.

Key concepts

Generalized Parallelogram Equality

The generalized parallelogram equality is a fundamental property in Hilbert spaces. It extends the familiar parallelogram law to sums of multiple vectors. In mathematical terms, if you have vectors \(x_1, x_2, \ldots, x_n\) in a Hilbert space \(H\), the equality is expressed as: \[ \sum_{\epsilon_i = \pm 1} \left\| \sum_{i=1}^{n} \varepsilon_i x_i \right\|^2 = 2^n \sum_{i=1}^{n} \left\| x_i \right\|^2 \] This means that summing the squared norms of all possible combinations of these vectors, each with plus or minus signs, equals \(2^n\) times the sum of the squared norms of the individual vectors. Understanding this property is essential for many applications in functional analysis and quantum mechanics. Let's explore how to prove this using mathematical induction.

Induction Proof

Proof by induction is a powerful technique in mathematics. It involves two main steps: establishing a base case and proving an induction step. Let's apply this to our generalized parallelogram equality.

• 1. **Base Case**: Start with \(n=1\). Verify that: \[ \sum_{\epsilon_1=\pm 1} \left\| \epsilon_1 x_1 \right\|^2 = 2 \left\| x_1 \right\|^2 \] This simply shows that for one vector, the equality holds as the norm squares of \(+x_1\) and \(-x_1\) sum to \(2 \left\| x_1 \right\|^2\).
• 2. **Induction Hypothesis**: Assume the equality holds for some \(n=k\). That is, \[ \sum_{\epsilon_i=\pm 1} \left\| \sum_{i=1}^{k} \varepsilon_i x_i \right\|^2 = 2^k \sum_{i=1}^{k} \left\| x_i \right\|^2 \]
• 3. **Induction Step**: Prove it for \(n=k+1\). Consider adding one more vector \(x_{k+1}\). By expanding the left-hand side using properties of inner product spaces and simplifying, you'll show that: \[ \sum_{\epsilon_i=\pm 1} \left\| \sum_{i=1}^{k+1} \varepsilon_i x_i \right\|^2 = 2^{k+1} \sum_{i=1}^{k+1} \left\| x_i \right\|^2 \]

This completes the induction proof, generalizing the parallelogram equality for all \(n\).

Inner Product Space

An inner product space is a vector space equipped with an operation called the inner product. The inner product \(\) of two vectors \(x\) and \(y\) returns a scalar and satisfies several properties:

• 1. **Linearity**: \(\forall x, y, z\) and scalars \(a, b\), \( = a + b \)
• 2. **Symmetry**: \( = \overline{}\), where \(\overline{<•, •>}\) denotes the complex conjugate.
• 3. **Positive-Definiteness**: \( \geq 0\) and \( = 0\) iff \(x = 0\).

These properties ensure that many familiar geometric concepts like angles, lengths, and orthogonality carry over to the abstract setting. In a Hilbert space—a complete inner product space—these concepts provide crucial tools for analysis in functional spaces.

Pythagorean Theorem

The Pythagorean theorem is one of the simplest yet most important results in geometry. It states that for a right-angled triangle with sides \(a\), \(b\), and hypotenuse \(c\): \[ a^2 + b^2 = c^2 \] In the context of inner product spaces, the Pythagorean theorem can be extended using the concept of orthogonality: \[ \| x + y \|^2 = \| x \|^2 + \| y \|^2 \] If vectors \(x\) and \(y\) are orthogonal, i.e., \( = 0\). This extends naturally in Hilbert spaces, providing a foundational tool for proving results like the generalized parallelogram equality. Orthogonality and the Pythagorean theorem help to simplify complex problems by breaking them into smaller, manageable orthogonal components.