Some musings on Cauchy-Schwarz, Part IVb
In this part we will look at two-dimensional real inner product spaces, which are all essentially identical to $\R^2$ with the standard dot product. They are also all essentially identical to $\C$ when $\C$ is regarded as a real inner product space, using the real inner product (as discussed in Part II).
\[\langle z,w \rangle = \Re(\bar z w)= \Re(z) \Re(w) + \Im(z) \Im(w)\,,\]
and the Cauchy-Schwarz inequality for all two-dimensional real inner product spaces is immediate from this, because $|\Re(\bar z w)| \leq |\bar z w|$. (Of course, Cauchy-Schwarz in $\R^2$ is easy to prove anyway!)
We already discussed the connection between $\R^2$ and $\C$ in Part II. Recall the following from that part (also discussed in intermediate parts).
The (linear in the second variable) standard complex inner product on $\C$ is given by $\langle z,w \rangle = \bar z w$, with the associated norm given by modulus. Writing $z=a_1+\i a_2$ and $w=b_1+\i b_2$, with $a_1,a_2,b_1,b_2$ real, we have $\Re(\bar z w)= a_1 b_1 + a_2 b_2$ which is just the usual dot product $(a_1,a_2)\cdot(b_1,b_2)$. So if we identify $\C$ with $\R^2$ in the usual way, then the usual dot product (on $\R^2$) is the real part of the usual complex inner product on the (one-dimensional) complex vector space $\C$.
The identifying bijection from $\R^2$ to $\C$ here is the map $(x,y)\mapsto x+\i y$ (which may actually be the identity map, depending on your construction of $\C$). This bijection is a real-linear map which preserves norms and the "usual" real inner products (the dot product on $\R^2$ and the real part of the usual complex inner product on $\C$.
As noted earlier, the Cauchy-Schwarz inequality for these two (identified) 2-dimensional real inner product spaces is immediate from the Cauchy-schwarz inequality for the 1-dimensional complex inner product space $\C$ (where equality holds), because $|\Re(\bar z w)| \leq |z||w|$. Of course, for $\R^2$, the Cauchy-Schwarz inequality is also immediate from the standard alternative formula for the dot product in $\R^2$ (and in $\R^3$), \[\va\cdot \vb = \|\va\| \|\vb\| \cos \theta,\]
where $\theta$ is the angle between $\va$ and $\vb$.
In Part IVc we will look properly at the claim that "all two-dimensional real inner product spaces are essentially identical to $\R^2$ with the standard dot product". This is very similar to the discussion in Part III, but needs just one more step in the Gram-Schmidt process. The same reasoning works for $n$-dimensional, real inner product spaces, which are all essentially the same as $\R^n$, and (with more theory and a bit more work) for separable, infinite-dimensional Hilbert spaces, which are all essentially the same as $\ell^2$. However my first target is just to understand the general two-dimensional real inner product spaces, then to deduce the general real Cauchy-Schwarz inequality from the two-dimensional case, and finally to deduce the Cauchy-Schwarz inequality for general complex inner product spaces.
Why all this fuss about a standard inequality for inner product spaces which has some very quick and elegant proofs available? Well, I just wanted to make it very clear that the basic Cauchy-Schwarz inequality really is a 2-dimensional result. This isn't the end of the story, though, as I hinted in Part I, when I asked the following.
Which came first, the chicken ... er, the Cauchy-Schwarz inequality or the triangle inequality? In many of the infinite-dimensional examples of inner product spaces in the Linear Analysis module (e.g. \(\ell^2\)) you need to verify that you really have a vector space (mainly checking that the space is closed under addition), and that the inner product is well-defined. Of course, once you know you have an inner product space, the Cauchy-Schwarz inequality is automatically true. But typically you need at least some "feeble versions" of the triangle inequality and the Cauchy-Schwarz inequality in order to check that you have an inner product space in the first place.
I plan to look at that in later parts. And then maybe look at the inequalities of Holder and Minkowski, if I have some more thoughts on those.
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