This blog post is entirely based on the start of this blog post by Terry Tao. I highly recommend reading the post. It gives an interesting insight into how Terry sometimes thinks about proving inequalities. He also gives a number of cool and more substantial examples.
The main idea in the blog post is that Terry likes to do “arbitrage” on an inequality to improve it. By starting with a weak inequality he exploits the symmetry of the environment he is working in to get better and better inequalities. He first illustrates this with a proof of the Cauchy-Schwarz inequality. The proof given is really nifty and much more memorable than previous proofs I’ve seen. I felt that just had to write it up and share it.
Let \((V,\langle, \rangle)\) be an inner product space. The Cauchy-Schwarz inequality states that for all \(v,w \in V\), \(|\langle v, w \rangle | \le \|v\| \|w\|\). It’s an important result that leads, among other things, to a proof that \(\| \cdot \|\) satisfies the triangle inequality. There are many proofs of the Cauchy-Schwarz inequality but here is the one Terry presents.
Since \(\langle \cdot, \cdot \rangle \) is positive definite we have \(0 \le \langle v-w,v-w\rangle\). Now using the fact that \(\langle \cdot, \cdot \rangle\) is additive in each coordinate we have
\(0 \le \langle v,v \rangle -\langle v, w \rangle -\langle w,v\rangle+ \langle w,w \rangle\).
Since \(\langle w,v\rangle = \overline{\langle v,w\rangle}\), we can rearrange the above expression to get the inequality
\(\text{Re}(\langle v,w \rangle) \le \frac{1}{2}\left(\| v \|^2 + \|w\|^2\right)\).
And now it is time to exploit the symmetry of the above expression and turn this inequality into the Cauchy-Schwarz inequality. The above inequality is worse than the Cauchy Schwarz inequality for two reasons. Firstly, unless \(\langle v, w \rangle\) is a positive real number, the left hand side is smaller than \(|\langle v,w \rangle|\). Secondly, unless \(\|v\|=\|w\|\), the right hand side is larger than the quantity \(\|v\|\|w\|\) that we want. Indeed we want the geometric mean of \(\|v\|^2\) and \(\|w\|^2\) whereas we currently have the arithmetic mean on the right.
Note that the right hand side is invariant under the symmetry \(v \mapsto e^{i \theta} v\) for any real number \(\theta\). Thus choose \(\theta\) to be the negative of the argument of \(\langle v,w \rangle\). This turns the left hand side into \(|\langle v,w \rangle |\) while the right hand side remains invariant. Thus we have done our first bit of arbitrage and now have the improved inequality
\(| \langle v,w \rangle | \le \frac{1}{2}\left(\|v\|^2 + \|w\|^2\right)\)
We now turn our attention to the right hand side and observe that the left hand side is invariant under the map \((v,w) \mapsto \left(c\cdot v, \frac{1}{c} \cdot w\right)\) for any \(c > 0\). Thus by choosing \(c\) we can minimize the right hand side. A little bit of calculus shows that the best choice is \(c = \sqrt{\frac{\|w\|}{\|v\|}}\) (this is valid provided that \(v,w \neq 0\), the case when \(v=0\) or \(w=0\) is easy since we would then have \(\langle v,w \rangle=0\)). If we substitute in this optimal value of \(c\), the right hand side of the above inequality becomes
\( \frac{1}{2}\left( \left\| \sqrt{\frac{\|w\|}{\|v\|}}\cdot v \right \|^2 +\left \| \sqrt{\frac{\|v\|}{\|w\|}} \cdot w\right \|^2 \right)=\frac{1}{2}\left(\frac{\|w\|}{\|v\|}\|v\|^2+\frac{\|v\|}{\|w\|}\|w\|^2 \right) = \|v\|\|w\|.\)
Thus we have turned our weak starting inequality into the Cauchy-Schwarz inequality! Again I recommend reading Terry’s original post to see many more examples of this sort of arbitrage and symmetry exploitation.
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