Question

Suppose \(\left\\{x^{k}\right\\}_{k=1}^{\infty}\) is an orthonormal sequence in \(\ell_{2}\), where \(x^{k}=\left(x_{i}^{k}\right)\). Show that \(\lim _{k \rightarrow \infty}\left(x_{i}^{k}\right)=0\) for every \(i \in \mathbf{N}\). Hint: Use the Bessel inequality to show that \(\left(e_{i}, x^{k}\right) \rightarrow 0\) as \(k \rightarrow \infty\).

Short answer

Using Bessel’s Inequality on \( e_{i} \), we show \( \backslash lim_{k \backslash rightarrow \backslash infty} x_{i}^{k} = 0 \) for every \(i\).

Step-by-step solution

Understand the Problem

Given an orthonormal sequence \(\backslash left\{x^{k}\right\}_{k=1}^{infty}\) in \(\backslash ell_{2}\), we need to show that \( \backslash lim\backslash _{k \backslash rightarrow \backslash infty}\backslash left(x\backslash _{i}^{k}\backslash right)=0\) for every \(\backslash i \backslash in \backslash mathbf{N}\).

Recall the Orthonormal Sequence Definition

A sequence \( \backslash left\{x^{k}\right\}\) is orthonormal in \(\ell_{2}\) if each element of the sequence is orthogonal to each other and each element has a norm of 1. This implies \( \backslash left\|x^{k}\right\|\backslash _{2} = 1 \) and \( \backslash left(e_{i}, x^{k}\right) = \backslash sum \backslash _{i=1}\backslash ^{\backslash infty} x^{i}y^{i} \backslash = 0 \) whenever \( i eq j \).

Use the Hint and Bessel's Inequality

According to Bessel's inequality, for any orthonormal sequence \( \backslash left\{x^{k}\right\}\) in \(\backslash ell_{2}\) and any vector \(y \backslash in \backslash ell_{2}\), we have \( \backslash sum \backslash _{k=1}\backslash ^{\backslash infty}\backslash |(y, x^{k})\backslash |^{2} \backslash \backslash leq \backslash backslash left\|y\right\|\backslash _{2}^{2}\).\Let us choose \(y = e_{i}\), the \(i\)-th standard basis vector in \(\ell_{2}\). Then \( (e_{i}, x^{k}) = x_{i}^{k} \). Therefore, Bessel's inequality gives us \( \backslash sum \backslash _{k=1}\backslash ^{\backslash infty}\backslash |x_{i}^{k}|^{2} \backslash \backslash leq 1\). This implies that the series \( \backslash left\|x_{i}^{k}\right\|\backslash _{2}^{2}\) converges and thus \( x_{i}^{k}\) must go to zero as \( k\) goes to infinity.

Conclusion

Since the sequence \( x_{i}^{k} \) is a Cauchy sequence due to the orthonormality and Bessel's Inequality, and every orthonormal sequence in \(\backslash ell_{2}\) converges in \( \backslash ell_{2}\), hence it must converge to zero. Therefore, \( \backslash lim\backslash _{k\rightarrow\backslash infty}x\backslash _{i}^{k}=0\) for every \(i \backslash in N\).

Key concepts

orthonormal sequence

An orthonormal sequence is a powerful concept in functional analysis and Hilbert space theory. It consists of vectors that are both orthogonal (perpendicular) to each other and normalized to have a unit length.
To understand this better, let’s break it down:

• Orthogonality: Two vectors are orthogonal if their inner product is zero. For vectors \(x\) and \(y\), this is written as \( \langle x, y \rangle = 0 \).
• Normalization: Each vector in the sequence has a norm (or length) of one. In mathematical terms, for vector \(x^k\), we have \( \|x^k\|_2 = 1 \).

This means that if we have a sequence of orthonormal vectors \( \{x^k\}_{k=1}^{\infty} \) in \( \ell_2\) space, they all sit at a mutual perpendicular distance from each other and each has a magnitude of 1.
Orthonormal sequences play a vital role in simplifying complex problems as they provide a neatly decomposed and independent set of vectors.

Bessel's inequality

Bessel’s inequality is a fundamental result in the realm of orthonormal sequences. It provides a bound on the sum of squares of coefficients when projecting a vector onto the elements of an orthonormal sequence.
Formally, let's say we have an orthonormal sequence \( \{x^k\}_{k=1}^{\infty} \) in \( \ell_2\) space and a vector \( y \) in the same space. Bessel’s inequality states:
\[ \sum_{k=1}^{\infty} \left| \langle y, x^k \rangle \right|^2 \leq \|y\|_2^2 \]This means that the sum of the squared coefficients—when we project \( y \) onto our orthonormal sequence—can never exceed the squared norm of \( y \).
In our exercise, we set \( y = e_i \), the \( i \)-th standard basis vector. This simplification gives us the relation:
\[ \sum_{k=1}^{\infty} \left| \langle e_i, x^k \rangle \right|^2 \leq 1 \]Since \( \left| \langle e_i, x^k \rangle \right| = \left|x_i^k\right| \), it implies that the series \( \sum_{k=1}^{\infty} \left|x_i^k\right|^2 \) converges and thus \( x_i^k \) approaches zero as \( k \) approaches infinity.

convergence in l2 space

Convergence in \( \ell_2\) space refers to the behavior of sequences whose elements are square-summable, meaning their squared norm converges to a finite value.
Let's say \( \{a_n\} \) is a sequence in \( \ell_2\) space. We say \( a_n \) converges in \( \ell_2\) if:
\[ \lim_{n \rightarrow \infty} \left\|a_n - a\right\|_2 = 0 \]where \( a \) is the limit of the sequence.
For sequences in \( \ell_2\), this means that the sum of the squared differences between the sequence elements and the limit goes to zero. This is crucial as it ensures the sequence 'stabilizes' and does not deviate wildly as it progresses.In our exercise, we are dealing with sequence elements represented as \( x_i^k \). By applying Bessel’s inequality, we showed that \( x_i^k \) goes to zero for every \( i\). Since these elements belong to an orthonormal sequence, this guarantees their convergence in \( \ell_2\).So, \:

• We can trust that the sequence elements settle down to zero.
• This aligns with the property that every orthonormal sequence in \( \ell_2\) converges in \( \ell_2\) space.

Hence, understanding convergence in \( \ell_2\) space helps in comprehending how sequences behave and ensures stability in their long-term predictions.