Prove inner-product inequality $\Bigl (\sum_{i=1}^k \langle u\;,\;e_i \rangle \langle v\;,\;e_i \rangle \Bigr)^2≤ \Vert u \Vert^2 \, \Vert v \Vert^2$

Question

The Question:

For the real vector space $V$, let $u,v \in V$ and $\{ e_1,\dots,e_k \}$ being a set of orthonormal vectors wrt to the inner product $\langle \cdot \;,\;\cdot \rangle$. Prove that

$$\biggl (\sum_{i=1}^k \langle u\;,\;e_i \rangle \langle v\;,\;e_i\rangle \biggr)^2≤ \Vert u \Vert^2 \, \Vert v \Vert^2$$

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My Attempt:

$\{ e_1,\dots,e_k \}$ being orthonormal means that we can extend it to an orthonormal basis $\{ e_1,\dots,e_n \}$ for $V$ (where $n≥k$ of course).

So let $u=\sum_{i=1}^n u_i e_i$ and $v=\sum_{i=1}^n v_i e_i$ for some $u_1,\dots ,u_n,v_1, \dots ,v_n \in \Bbb R$. Then

\begin{align} \biggl (\sum_{i=1}^k \langle u\;,\;e_i \rangle \langle v\;,\;e_i \rangle \biggr)^2 & = \biggl (\sum_{i=1}^k \langle \sum_{j=1}^n u_j e_j\;,\;e_i \rangle \langle \sum_{j=1}^n v_j e_j\;,\;e_i \rangle \biggr)^2 \\ & = \biggl (\sum_{i=1}^ku_iv_i \biggr)^2 \end{align}

and I am stuck.

If only I could show that

$$\biggl (\sum_{i=1}^ku_iv_i \biggr)^2≤\biggl (\sum_{i=1}^nu_iv_i \biggr)^2$$

then using Cauchy-Schwarz Inequality would get us there directly.

Any hints?

Answer 1

Your assertion is false. As an example take $k=1$, $n=2$ and $u=(2,-1)$,$v=(1,1)$, which gives you $2^2\leq (2-1)^2$, i.e. false.

Instead, you can apply CS directly on the $k$ sum, giving:

$$(\sum_{i=1}^ku_iv_i)^2\leq \|u(1:k)\|^2\|v(1:k)\|^2,$$

where the 1:k notation means, the first $k$ components. Now just easily prove $\|u(1:k)\|^2\leq \|u\|^2$, and similarly for $v$.