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From Munkres' Topology, I get this question. A hint suggests us to use a result from a previous subquestion. But it seems that I don't need to use the previous result to prove this. Can someone help me to see if I can do that? The picture is the exercise that I'm working on. Following that is a proof for $(c)$. enter image description here

Proof: $$\begin{align*} ||x+y||^2 &= (x+y)(x+y)\\ &=(x+y)x+(x+y)y\\ &=xx+yx+xy+yy\\ &=||x||^2+2(xy)+||y||^2\\ & \leq ||x||^2+2||x||||y||+||y||^2\\ &=(||x||+||y||)^2. \end{align*}$$ Since $||x+y||$ and $(||x||+||y||)$ in the set of nonnegative real numbers, we have $$||x+y||\leq||x||+||y||.$$

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  • $\begingroup$ Yep (once you fix the squaring). As appears in innumerable books and online. $\endgroup$ – David G. Stork Mar 14 '18 at 20:48
  • $\begingroup$ @DavidG.Stork But why did Munkres give a hint to use the result $(b)$? What's his intention? $\endgroup$ – user398843 Mar 14 '18 at 20:50
  • $\begingroup$ Since $x, y \in \mathbb{R}^n$ are not single numbers, you need to use the dot product in the expressions, and then you would be using (b) in the inequality from line 4 to line 5. $\endgroup$ – Daniel Schepler Mar 14 '18 at 20:51
  • $\begingroup$ @BarryCipra Thank you for pointing that out. $\endgroup$ – user398843 Mar 14 '18 at 20:51
  • $\begingroup$ @user398843, you used (b) in going from the fourth to the fifth line (where the $\le$ sign appears). $\endgroup$ – Barry Cipra Mar 14 '18 at 20:52

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