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I'm reading Steele's: The Cauchy-Schwarz Master Class.


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I'm having some trouble understanding it, I'll list my doubts:

  1. I'm having trouble understanding $\sum_{k=1}^{\infty}a^2_k<\infty$. It's lesser than infinity? What does that mean? In the following text, it mentions that it is small but I'm not sure of what that means, small in relation to what?

  2. I don't know how he goes from $(1.4)$ to $xy\leq C(x^2+y^2)$.

  3. I don't know how he draws the link to the familiar factorization. I can only think of $xy\leq C(x^2+y^2)$ as $0\leq Cx^2+Cy^2-xy$. I see that $xy\leq C(x^2+y^2)$ has a connection with $(1.5)$ in which it seems he's assuming that $C=1/2$. I'm lost at this mid step.

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    $\begingroup$ $\lt\infty$ usually means convergence/finiteness. $\endgroup$ – Ian Mateus Dec 19 '13 at 22:14
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    $\begingroup$ The second part means that if you are able to show $a_kb_k\leqslant C(x_k^2+y_k^2)$, then $\sum a_kb_k\leqslant C\left(\sum a_k^2+\sum b_k^2\right)$, implying 1.4. $\endgroup$ – Ian Mateus Dec 19 '13 at 22:22
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1

When working with the non-negative reals, it is often useful to treat $\infty$ as a formal symbol which is bigger than everything else (see extended reals). Then a sum of non-negative numbers always has a value; either it's infinite, or it converges to a real number (this statement is related to the completeness of the reals, and is worth thinking about if you haven't already).

So, writing $\sum_{k=1}^\infty a_k^2 < \infty$ just means that the series $\sum_{k=1}^\infty a_k^2$ converges to some non-negative real number. Equivalently, it means that all the finite sums $\sum_{k=1}^N a_k^2$ are bounded above by some fixed number. The least upper bound is what we call the "value" of the series $\sum_{k=1}^\infty a_k^2$.

For example, we could write $\sum_{k=1}^\infty \frac{1}{2^k} <\infty$, because $\sum_{k=1}^N\frac{1}{2^k} = 1 - \frac{1}{2^N}$, so the finite sums are bounded above by $1$. In fact, $\sum_{k=1}^\infty \frac{1}{2^k} =1$.

2

It's important to realize that the author is basically thinking backwards here. That is, he's saying: What would it take to prove (1.4)? Well, we know that two things are finite, and we want to prove that a third thing is finite. One way to do this would be to show that it's less than a constant times the sum of the first two things.

In other words, if we knew that there was some non-negative $C$ with $|a_k b_k| \leq C(a_k^2 + b_k^2)$ for all $k$, then that would be enough to prove (1.4), because then we would have $\sum_{k=1}^N |a_k b_k| \leq C (\sum_{k=1}^N a_k^2 + \sum_{k=1}^N b_k^2)$. If the two sums on the right are bounded above by non-negative reals $A$ and $B$, then it follows that the sum on the left is bounded above by the non-negative real $C(A+B)$.

Since there is nothing particularly special about the numbers $|a_k|$ and $|b_k|$, the author is forgetting about the sequences for the moment and considering any two non-negative real numbers $x$ and $y$. If it were true that there was some $C$ with $xy \leq C(x^2 + y^2)$ for any non-negative $x$ and $y$, then we could set $x=|a_k|$ and $y=|b_k|$ and get exactly the thing we wanted.

(Note that we haven't really done any math yet; this is a style of reasoning I like to call "wishful thinking", where you imagine what would be nice to have, or allow you to solve the problem, but you don't immediately worry about whether it's possible or not. Here we are very lucky, and we will get everything that we hope for, but for the moment we are just asking "What if?")

3

Again, you need to understand that the author is thinking backwards. We don't know that $xy\leq C(x^2+y^2)$, this is something that we hope is true for some $C$. So there is no problem trying out $C=\frac{1}{2}$; if that works, then we have what we want.

But why would we guess that number? Well, maybe we're staring at the equation $Cx^2 + Cy^2 - xy \geq 0$ and we realize that it looks a little bit like a square. Maybe we even try multiplying by $2$, to get $2Cx^2 + 2Cy^2 - 2xy \geq 0$. Now it looks a whole lot like $(x-y)^2 \geq 0$, which we already know is true! So the thing that want to be true really is true when $C=\frac{1}{2}$—which is very nice for us.

But the reasoning goes in the opposite direction of the writing. That is, we know that $(|a_k| - |b_k|)^2 \geq 0$, therefore $|a_k b_k| \leq C (a_k^2 + b_k^2)$ for some non-negative $C$, therefore we can bound the series $\sum_k |a_k b_k|$, etc.

The author is not giving the proof; he is telling you the right way to think about a problem like this. You cannot always directly deduce the thing you want; sometimes you need to begin with the conclusion, and ask what steps would lead you there. There are no fixed methods for doing this. It is just asking the question "What if?" For any given problem, this approach may not work, or it may work only partially, but here it leads to a solution very, very quickly.

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  1. $\lt\infty$ means finiteness in this context.

  2. You want to show that $\sum a_kb_k$ converges given $\sum a_k^2$ and $\sum b_k^2$. So what can you do? The strategy the author uses is to bound the sum by $C\left(\sum a_k^2+\sum b_k^2\right)$, so we can show it is finite.

  3. The connection you describe can be seen essentially in at least two ways: first, the $\text{AM-GM}$, and second, by its proofs. The easiest proof I know is exactly the one the author makes use: $(x-y)^2\geqslant 0$, so $2xy\leqslant x^2+y^2$, implying the claim. One way, then, is to be familiar with $\text{AM-GM}$; knowing it and its proof, you can connect things here.
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