# Thread: A question on normed vector space

1. ## A question on normed vector space

This is from problem 1.1 of the book "Analysis for Applied Mathematics" written by Ward Cheney.

17. Prove that in a normed linear space, if $\left \| x+y \right \| = \left \| x\right \| + \left \| y\right \|$ then $\left \| ax+by \right \| = \left \| ax\right \| + \left \| by\right \|$ for all nonnegative $a, b|$

If $\left \| x+y\right \| = \left \| x\right \| + \left \| y\right \|$ if and only if $x=ky$ for some scalar $k$, then this problem can be solved easily. But I guess this property is not generally hold.

I cannot find a counter example and cannot prove it too. Anyone has an idea?

2. ## Re: A question on normed vector space

Sorry, I made some errors on the post and I don't know how to edit it.
This is the edited version:

This is from problem 1.1 of the book "Analysis for Applied Mathematics" written by Ward Cheney.

17. Prove that in a normed linear space, if ||x+y|| = ||x|| + ||y|| then ||ax+by|| = ||ax|| + ||by|| for all nonnegative a, b

If ||x+y|| = ||x|| + ||y|| implies x = ky for some scalar k, then this problem can be solved easily. But I guess this property does not hold generally.

I cannot find a counter example and cannot prove it too. Anyone has an idea?

3. ## Re: A question on normed vector space

Originally Posted by dotalam
If ||x+y|| = ||x|| + ||y|| implies x = ky for some scalar k, then this problem can be solved easily. But I guess this property does not hold generally.
Indeed, if you consider $\mathbb R^2$ and $\lVert (x,y)\rVert:=|x|+|y|$ then we have $\lVert (1,1)+(1,0)\rVert =\lVert (1,1)\rVert +\lVert (1,0)\rVert$ and we can't find k such that $(1,1)=k(1,0)$.

4. ## Re: A question on normed vector space

Originally Posted by dotalam
Prove that in a normed linear space, if $\left \| x+y \right \| = \left \| x\right \| + \left \| y\right \|$ then $\left \| ax+by \right \| = \left \| ax\right \| + \left \| by\right \|$ for all nonnegative $a, b.$
If x or y is zero then the result is obvious. So assume that neither of those elements is zero. The elements $u = \tfrac x{\|x\|}$, $v = \tfrac y{\|y\|}$ and $w = \tfrac {x+y}{\|x+y\|}$ all have norm 1. But $w = \tfrac{\|x\|}{\|x\|+\|y\|}u + \tfrac{\|y\|}{\|x\|+\|y\|}v$, so w lies on the line segment from u to v. Use the convexity of the unit ball to deduce that every element on that line segment has norm 1. The result $\left \| ax+by \right \| = \left \| ax\right \| + \left \| by\right \|$ for all nonnegative $a, b$ should then follow.

LaTeX tip: In this forum, use the [TEX]..[/TEX] tags, not $$..$$.

5. ## Re: A question on normed vector space

Originally Posted by Opalg
Use the convexity of the unit ball to deduce that every element on that line segment has norm 1[/noparse]
How to deduce every element on that line segment has norm 1?? The direct implication of the convexity of the unit ball should be every element on that line segment has norm not greater than 1. Thank you!

6. ## Re: A question on normed vector space

Originally Posted by dotalam
How to deduce every element on that line segment has norm 1?? The direct implication of the convexity of the unit ball should be every element on that line segment has norm not greater than 1. Thank you!
Here's the geometric picture. It's up to you to translate it into an analytic argument.

Suppose that x, y, z are points all having norm 1, with z lying on the line segment from x to y. Let w be another point on that line segment, and suppose that $\|w\|<1.$ Let v be the positive-scalar multiple of w whose norm is equal to 1. The line segment from v to y must lie in the unit ball. But z lies strictly inside that line segment (in other words, between the line segment and the origin), which means that $\|z\|<1$ — contradiction. Therefore $\|w\|=1.$