In class (as is the case also in Anton’s book) the strategy for establishing the triangle inequality ||u+v||<||u|| + ||v|| was to establish first the Cauchy-Schwartz Inequality and then to argue as follows, given vectorsuandvin someR^{n}(orR_{n}):

1. Expand and then estimate ||u+v||^{2}:|| 2. take square roots on both sides:u+v||^{2}= (u+v) • (u+v) = (u•u) + 2(u•v) + (v•v)

<(u•u) + 2 |u•v| + (v•v) = ||u||^{2}+ 2 |u•v| + ||v||^{2}

<||u||^{2}+ 2 ||u|| ||v|| + ||v||^{2}= (||u|| + ||v||)^{2}; and then|| u+v||<||u|| + ||v|| .

Amusingly enough, one can also reverse this inference: that is, it is possible to derive the Cauchy-Schwartz inequality from the triangle inequality; here’s how.

Recall (from class,^{ }or from Anton, Chapter 4, section 1, item 4.1.6) the formulathough it will u•v= ¼[(||u+v||)^{2}– (||u–v||)^{2}] ,^{ }only get used at the very end, after we’ve prepared suitable estimates on the two norms (squared) within the RHS.

On the one hand,^{ }applying the triangle inequality tou–vandvtells us ||u|| = ||(u–v) +v||<||u–v|| + ||v|| , which rebalances to|| and applying it instead tou|| – ||v||<||u–v|| ;v–uandutells us ||v|| = ||(v–u) +u||<||v–u|| + ||u|| , which rebalances and rewrites to|| Butv|| – ||u||<||v–u|| = ||u–v|| .^{ }these two inequalities taken together can be summarized as ||u–v||>| ||u|| – ||v|| | , whence|| On the other hand,u–v||^{2}>(||u|| – ||v||)^{2}.^{ }the triangle inequality applied touandvtells us ||u+v||<||u|| + ||v|| , whence|| From these last two inequalities it follows thatu+v||^{2}<(||u|| + ||v||)^{2}.|| andu+v||^{2}– ||u–v||^{2}<(||u|| + ||v||)^{2}– (||u|| – ||v||)^{2}= ... = 4 ||u|| ||v|| (why?),^{ }if we apply the thinking above not touandvbut touand –v, we find (since || –v|| = ||v|| ):|| Finally,u–v||^{2}– ||u+v||^{2}<(||u|| + ||v||)^{2}– (||u|| – ||v||)^{2}= 4 ||u|| ||v|| .^{ }these two inequalities taken together tell us that| || and so,u+v||^{2}– ||u–v||^{2}|<4 ||u|| ||v|| ,^{ }at last exploiting the formula foru•vdisplayed at the outset, we have| as was to be proved.u•v| = | ¼[||u+v||^{2}– ||u–v||^{2}] | = ¼ | ||u+v||^{2}– ||u–v||^{2}|<¼ (4 ||u|| ||v||) = ||u|| ||v|| ,

First posted: 2 Apr 2006. Math 221 Home Contents © 2006 by Fred E.J. Linton