Some musings on Cauchy-Schwarz, Part III

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In this part we look at the very trivial case of one-dimensional inner product spaces over $\R$. As an exercise you can check any details omitted below, and identify the famous named theorems we are discussing the one-dimensional versions of!

Almost everything discussed below is just the trivial one-dimensional case of standard facts about finite-dimensional, real inner product spaces.

More advanced readers may wish to skip or merely skim Parts III and IV and move on to Part V.

The standard inner product on $\R$ is just the usual product in $\R$, so, for $x,y \in \R$,

\[\langle x,y\rangle_\R=xy\,.\]

This is, of course, the one-dimensional version of the dot product!

You can check that this really is a (real) inner product. (This is mostly just the standard rules of arithmetic in $\R$, along with properties of modulus.)

Unsurprisingly, the associated norm is given by

\[||x||_\R=\sqrt{\langle x,x \rangle}_\R = \sqrt{x^2} = |x|\,,\]

i.e., the usual modulus of $x$.

In this special case, the Cauchy-Schwarz inequality claims that, for all $x,y \in \R$ we have 

\[|xy| \leq |x| |y|\,,\]

but of course we actually have equality here. So the inequality is clearly true here (remembering that equality does imply $\leq$).

Now let $V$ be any one-dimensional inner product space over $\R$, with inner product $\langle \cdot,\cdot \rangle_V$ and associated norm $\|\cdot\|_V$.

Let $\{\vv\}$ be a basis of $V$ (here $\vv$ can, of course, be any non-zero element of $V$). Then $\vv\neq 0$, so we normalize by setting

\[\Ve=\frac{1}{\|\vv\|_V} \vv\,.\]

Then $\{\Ve\}$ is (trivially) an orthonormal basis of $V$. (This is effectively the initial step of a Gram-Schmidt process, but it finishes already at this step.)

Define a map $T:\R\to V$ by $T(x)=x \Ve$. Then $T$ is easily seen to be a bijection. Moreover, $T$ is a linear isomorphism, and for all $x,y \in \R$, we have

\[\|T(x)\|_V=\|x\|_\R=|x|\]

and

\[\langle T(x),T(y)\rangle_V=\langle x,y\rangle_\R=xy\,.\]

In other words, $T$ preserves norms and inner products.

From this we see that all one-dimensional real inner product spaces are essentially the same as $\R$ with its standard inner product (which is just the usual product). Unsurprisingly, this means that equality always holds in the Cauchy-Schwarz inequality in one-dimensional real inner product spaces. You can check this directly, or use the "identification" above.

In the next post we look at the almost equally trivial case of two-dimensional real inner product spaces, and compare these with $\R^2$ and $\C$. (Some of this was discussed already in Part II.)

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