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Inner Product is a Continuous Mapping 📂Hilbert Space

Inner Product is a Continuous Mapping

Theorem1

Let’s assume (X,⟨⋅,⋅⟩)\left( X, \left\langle \cdot,\cdot \right\rangle \right) is an inner product space and {xn}\left\{ \mathbf{x}_{n} \right\}, {yn}\left\{ \mathbf{y}_{n} \right\} are sequences in XX converging to x\mathbf{x} and y\mathbf{y}, respectively. Then, the following holds.

⟨xn,yn⟩→⟨x,y⟩ as n→∞ \left\langle \mathbf{x}_{n},\mathbf{y}_{n} \right\rangle \to \left\langle \mathbf{x},\mathbf{y} \right\rangle \text{ as } n \to \infty

Since the limit can move inside and outside of the inner product, we obtain the following corollary.

Corollary

Assuming that HH is a Hilbert space, {xk}k∈N\left\{ \mathbf{x}_{k} \right\}_{k\in \N} is a sequence in HH, and {ck}k∈N∈ℓ2(N)\left\{ c_{k} \right\}_{k\in\mathbb{N}} \in \ell^{2} (\mathbb{N}) with ∑k=1∞ckxk\sum \limits_{k=1}^{\infty}c_{k}\mathbf{x}_{k} converging, then the following equation holds.

⟨x,∑k=1∞ckxk⟩= ∑k=1∞ck‾⟨x,xk⟩⟨∑k=1∞ckxk,x⟩= ∑k=1∞ck⟨xk,x⟩ \begin{align*} \left\langle \mathbf{x},\sum \limits_{k=1}^{\infty}c_{k}\mathbf{x}_{k} \right\rangle =&\ \sum \limits_{k=1}^{\infty}\overline{c_{k}}\left\langle \mathbf{x}, \mathbf{x}_{k} \right\rangle \\ \left\langle \sum \limits_{k=1}^{\infty}c_{k}\mathbf{x}_{k}, \mathbf{x} \right\rangle =&\ \sum \limits_{k=1}^{\infty}c_{k}\left\langle \mathbf{x}_{k}, \mathbf{x} \right\rangle \end{align*}

Explanation

lim⁡n→∞⟨xn,yn⟩=⟨lim⁡n→∞xn,lim⁡n→∞yn⟩=⟨x,y⟩ \lim \limits_{n\to \infty} \left\langle \mathbf{x}_{n}, \mathbf{y}_{n} \right\rangle = \left\langle \lim \limits_{n\to \infty} \mathbf{x}_{n}, \lim \limits_{n\to \infty} \mathbf{y}_{n} \right\rangle = \left\langle \mathbf{x},\mathbf{y} \right\rangle

This implies that the inner product is a continuous mapping, according to the equivalence condition of continuity. It is needless to say that this is a very useful property.

Proof

It can be easily shown through the definition of inner product and the Cauchy-Schwarz inequality.

∣⟨xn,yn⟩−⟨x,y⟩∣= ∣⟨xn,yn⟩−⟨x,yn⟩+⟨,yn⟩−⟨x,y⟩∣≤∣⟨xn,yn⟩−⟨x,yn⟩∣+∣⟨x,yn⟩−⟨x,y⟩∣= ∣⟨xn−x,yn⟩∣+∣⟨x,yn−y⟩∣ \begin{align*} \left| \left\langle \mathbf{x}_{n},\mathbf{y}_{n} \right\rangle -\left\langle \mathbf{x},\mathbf{y} \right\rangle \right| =&\ \left| \left\langle \mathbf{x}_{n},\mathbf{y}_{n} \right\rangle -\left\langle \mathbf{x},\mathbf{y}_{n} \right\rangle + \left\langle ,\mathbf{y}_{n} \right\rangle -\left\langle \mathbf{x},\mathbf{y} \right\rangle \right| \\ \le& \left| \left\langle \mathbf{x}_{n},\mathbf{y}_{n} \right\rangle -\left\langle \mathbf{x},\mathbf{y}_{n} \right\rangle \right| +\left| \left\langle \mathbf{x},\mathbf{y}_{n} \right\rangle -\left\langle \mathbf{x},\mathbf{y} \right\rangle \right| \\ =&\ \left| \left\langle \mathbf{x}_{n}-\mathbf{x},\mathbf{y}_{n} \right\rangle \right| +\left| \left\langle \mathbf{x},\mathbf{y}_{n}-\mathbf{y}\right\rangle \right| \end{align*}

Applying the Cauchy-Schwarz inequality to the right-hand side gives the following.

∣⟨xn,yn⟩−⟨x,y⟩∣≤∥xn−x∥∥yn∥+∥x∥∥yn−y∥ \left| \left\langle \mathbf{x}_{n},\mathbf{y}_{n} \right\rangle -\left\langle \mathbf{x},\mathbf{y} \right\rangle \right| \le \left\| \mathbf{x}_{n}-\mathbf{x} \right\| \left\| \mathbf{y}_{n} \right\| + \left\| \mathbf{x} \right\| \left\| \mathbf{y}_{n}-\mathbf{y} \right\|

Since the norm is a continuous mapping, lim⁡n→∞∥xn−x∥=0\lim \limits_{n\to \infty} \left\| \mathbf{x}_{n}-\mathbf{x} \right\| =0 and the same applies to yn\mathbf{y}_{n}. Therefore, taking the limits on both sides of the equation gives

lim⁡n→∞∣⟨xn,yn⟩−⟨x,y⟩∣≤0  ⟹  lim⁡n→∞⟨xn,yn⟩= ⟨x,y⟩ \begin{align*} && \lim \limits_{n\to \infty} \left| \left\langle \mathbf{x}_{n},\mathbf{y}_{n} \right\rangle -\left\langle \mathbf{x},\mathbf{y} \right\rangle \right| \le& 0 \\ \implies && \lim \limits_{n\to \infty} \left\langle \mathbf{x}_{n}, \mathbf{y}_{n} \right\rangle =&\ \left\langle \mathbf{x},\mathbf{y} \right\rangle \end{align*}

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  1. Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications (2nd Edition, 1999), p173 ↩︎