if $\|x+y\|=\|x\|+\|y\|$, then $\|\alpha x+\beta y\|=\alpha \|x\|+\beta \|y\|$

Question

Let $X$ be a normed linear space. Assume that for $x,y \in X$, we have $||x+y||=||x||+||y||$. Show that $||\alpha x+\beta y||=\alpha ||x||+\beta ||y||$ for every $\alpha,\beta \geq 0$.

My attempt: Suppose $\beta \geq \alpha$. Then $\|\alpha x+\beta y\|=\|\alpha (x+y)-(\alpha-\beta)y\| \leq |\alpha|(\|x\|+\|y\|)+|\beta - \alpha|\|y\|=\alpha\|x\|+\beta\|y\|$.

Since the inequality is true for all $\alpha,\beta \geq 0$ and $\beta \geq \alpha$, if we let $\alpha=\beta=1$, we have $||x+y|| \leq |x||+||y||$. If the inequality is a strict inequality, then this contradicts with $\|x+y\|=\|x\|+\|y\|$. Hence, we must have equality, which is the desired result.

Is my proof correct?

EDIT: Suppose we have $\|x+y\|=\|x\|+\|y\|$ for some $x,y \in X$. For the same $x$ and $y$, prove that $\| \alpha x+ \beta y\|=\alpha \|x\|+\beta \|y\|$

This is a question from the book:'Banach Space Theory: The Basis for Linear and Nonlinear Analysis'. I copy the question without altering anything. — Idonknow, Dec 18 '13 at 18:38
That doesn't sound convincing. You seem to be assuming that if the equality is strict for some $\alpha$, $\beta$, then it also has to be strict for $\alpha=\beta=1$, but where does that assumption come from. — hmakholm left over Monica, Dec 18 '13 at 18:39
Your argument is, shall I say, overly optimistic. The inequality could be an equality for $\alpha = \beta = 1$, and a strict inequality for some other values. — TonyK, Dec 18 '13 at 18:41

Stephen Montgomery-Smith · Accepted Answer · 2013-12-19T00:39:36.950

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W.l.o.g. $\alpha \ge \beta$.

$$ \alpha\|x+y\| = \|(\alpha-\beta)y + (\alpha x + \beta y) \| \le (\alpha-\beta)\|y\| + \|\alpha x + \beta y\| \le (\alpha-\beta)\|y\| + \alpha\|x\| + \beta \|y\| = \alpha(\|x\|+\|y\|) $$ Since the first-quantity = last-quantity, all the inequalities must be equalities. In particular: $$(\alpha-\beta)\|y\| + \|\alpha x + \beta y\| = \alpha(\|x\|+\|y\|) .$$

edited Dec 19 '13 at 00:39

answered Dec 19 '13 at 00:09

Stephen Montgomery-Smith

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What motivates you to consider this approach? – Idonknow Dec 19 '13 at 05:37
It really was just symbol manipulation. Maybe I had seen this many years ago, and was regurgitating some long submerged memories. – Stephen Montgomery-Smith Dec 19 '13 at 05:40

score 1 · Answer 2 · answered Dec 18 '13 at 20:45

I'm not sure what you mean by "normed linear space". The usual definition includes, for example, the requirement that $\| \alpha x \| = |\alpha| \| x \|$, which equals $\alpha \| x\|$ if $\alpha \geq 0$, from which the claim follows trivially.

Perhaps your definition does not have that condition. Any function $\|\cdot\|$ satisfying $\|x+y\| = \|x\| + \|y\|$ for all $x,y$ satisfies $\|mx\| = m\|x\|$ for $m\in \{1,2,\dots\}$ by induction. This implies that $\|rx\| = r\|x\|$ for $r$ a positive rational number, by division. If you know your function to be continuous, the equation for $r$ real follows.

Actually, note that if $\|x+y\| = \|x\| + \|y\|$ for all $x,y\in X$, then $\|x\| = \|x+0\| = \|x\| + \|0\|$, and so by subtracting $\|0\| = 0$. Then $0 = \|0\| = \|x + (-x)\| = \|x\| + \|-x\|$, and so $\|-x\| = -\|x\|$, and the formula $\|rx\| = r\|x\|$ follows for all (possibly negative) rational $r$, and all real $r$ if $\|\cdot\|$ is continuous.

However, the usual definition of "norm" requires instead that $\|rx\| = |r|\|x\|$. For $r$ negative, we conclude that $r\|x\| = -r\|x\|$, from which it follows that $\|x\| = 0$. If your definition includes the condition $\forall x((\|x\| = 0) \Rightarrow (x = 0))$, then your vector space $X$ is the zero vector space.

So for the problem to be nontrivial, my conclusion is that either you are using funny definitions, or leaving out some quantifiers.

If think that the question meant: Suppose there exists $x$, $y$, $X$, where $X$ is a normed linear space, and $x,y\in X$, we have $|x+y|=|x|+|y|$. Then for that specific $x$ and $y$, show that we have the equality: for all $\alpha,\beta \ge 0$, we have $|\alpha x+\beta y| = \alpha|x| + \beta|y|$. — Stephen Montgomery-Smith, Dec 18 '13 at 23:56

Raul C. de Assis · Answer 3 · 2013-12-19T00:55:03.000

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I believe this can be answered in a much more basic way.

If X is a normed linear space, given $x,y \in X$ we have that $\alpha x, \beta y \in X$, because $X$ is closed under scalar multiplication.

So if $||x+y|| = ||x|| + ||y||$ $\forall x,y \in X$ we must also have $||\alpha x + \beta y|| = ||\alpha x|| + ||\beta y||= |\alpha|||x|| + |\beta|||y||= \alpha||x|| + \beta||y||$

Since $\alpha$ and $\beta$ $\geq0$.

edited Dec 19 '13 at 00:55

answered Dec 19 '13 at 00:25

Raul C. de Assis

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Look at the comment I made to Theo Johnson-Freyd's answer. If the OP had meant $|x+y| = |x|+|y|$ for all $x,y \in X$, we could deduce that $X$ is zero-dimensional, otherwise $y = -x$ would be a counterexample. – Stephen Montgomery-Smith Dec 19 '13 at 00:35
Maybe you are right and I misunderstood the question. You are right that it would be the linear space {0}. But even though if for any fixed $x_0,y_0 \in X$ and $X$ is a linear space we must have $\alpha x_0, \alpha y_0 \in X$ because $X$ is still closed under scalar multipliction, and the proof would follow the same way. – Raul C. de Assis Dec 19 '13 at 00:41
OK. But how do you go from $|\alpha x| + |\alpha y| = |\alpha||x| + |\beta||y|$? – Stephen Montgomery-Smith Dec 19 '13 at 00:43
When you define a norm ($||\cdot||$) on a linear space $X$ one of the properties of the norm is that $||\alpha x|| = |\alpha|||x||$, for all $x \in X$ and for all $\alpha$ in the respective scalar field of the linear space, in this case $\mathbb{R}$.
http://en.wikipedia.org/wiki/Norm_(mathematics)
– Raul C. de Assis Dec 19 '13 at 00:45
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Yes, but you changed an $\alpha$ on the LHS to a $\beta$ on the RHS. – Stephen Montgomery-Smith Dec 19 '13 at 00:47
I haven't realized, it was a typing mistake. – Raul C. de Assis Dec 19 '13 at 00:54
But now my original objection to your proof still stands. – Stephen Montgomery-Smith Dec 19 '13 at 00:56
let us continue this discussion in chat – Raul C. de Assis Dec 19 '13 at 01:00

if $\|x+y\|=\|x\|+\|y\|$, then $\|\alpha x+\beta y\|=\alpha \|x\|+\beta \|y\|$

3 Answers3