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I'm working through Spivak's Calculus textbook on Chapter 1, 19d.

The question posed is as follows:

FIRST IMAGE

SECOND IMAGE

Now the part I'm lost on is 19d. I was able to find the only when cases for parts a and b, but now I'm lost for part c. How do I find the case that equality holds only when $x_1 = (lambda)(y_1)$ and $x_2 = (lambda)(y_2)$?

A hint is fine, I'm not just looking for an answer. I'm just really stuck on this one. I have a feeling it has something to do with the terms $(x_1y_1 + x_2y_2)^2$ or $(x_2y_1 - x_1y_2)^2$ but I'm not sure.

Thank you ahead of time.

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For the (c) part of 19(d), given that equality of Schwarz inequality holds, i.e.

$$\color{red}{(x_1y_1 + x_2y_2)^2} = \color{blue}{(x_1^2+x_2^2)(y_1^2+y_2^2)}$$

Using $\color{blue}{(x_1^2+x_2^2) (y_1^2+y_2^2)} = \color{red}{(x_1y_1 + x_2y_2)^2} + \color{green}{(x_1y_2 - x_2y_1)^2}$, the identity from part (c):

$$\begin{align*} \color{green}{(x_1y_2-x_2y_1)^2} &= 0\\ x_1y_2 - x_2y_1 &= 0\\ x_1y_2 &= x_2y_1\\ \end{align*}$$

Either $y_1=y_2=0$, or WLOG assume $y_2 \ne 0$,

$$\begin{align*} x_1 &=\frac{x_2y_1}{y_2}\\ x_1 &= \frac{x_2}{y_2} \cdot y_1 \end{align*}$$

By taking $\lambda = \frac{x_2}{y_2}$, there is a number $\lambda$ that satisfies both $x_1 = \lambda y_1$ and $x_2 = \lambda y_2$.

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  • $\begingroup$ But why does part c tell us that $(x_1y_2 - x_2y_1)^2 = 0$ for equality to hold? Because equality holds only when $(x_1^2 + x_2^2)(y_1^2 + y_2^2) ⩾ (x_1)(y_1) + (x_2)(y_2)$ ? $\endgroup$ – psa Sep 29 '18 at 23:52
  • $\begingroup$ @kyle: square both sides of the Schwarz inequality (which we’re assuming is an equality here). This means the entire left-hand side equals the first term on the right-hand side. This, the second term must be equal to $0$. $\endgroup$ – Clayton Sep 29 '18 at 23:57
  • $\begingroup$ @kylecampbell Equality holds implies (by squaring both sides of Schwarz inequality) $$\color{red}{(x_1y_1 + x_2y_2)^2} = \color{blue}{(x_1^2+x_2^2)(y_1^2+y_2^2)}$$ Part (c) says $$\color{blue}{(x_1^2+x_2^2) (y_1^2+y_2^2)} = \color{red}{(x_1y_1 + x_2y_2)^2} + \color{green}{(x_1y_2 - x_2y_1)^2}$$ Then $\color{green}{(x_1y_2 - x_2y_1)^2}$ can only be $0$. $\endgroup$ – peterwhy Sep 30 '18 at 0:00
  • $\begingroup$ @kylecampbell I assumed you directly used the identity in part (c) and the inequality $$(x_1y_2 - x_2y_1)^2 \ge 0 \tag 1$$ to prove Schwarz inequality. If equality holds for Schwarz inequality, then equality should also hold for inequality $(1)$. $\endgroup$ – peterwhy Sep 30 '18 at 0:06
  • $\begingroup$ I see now. Thanks a lot Clayton and @peterwhy $\endgroup$ – psa Sep 30 '18 at 0:09

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