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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 $\| \alpha x+ \beta y\|=\alpha \|x\|+\beta \|y\|$

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  • $\begingroup$ ' Assume that for $x,y∈X$' quantify $x$ and $y$, please. $\endgroup$ – Git Gud Dec 18 '13 at 18:35
  • $\begingroup$ This is a question from the book:'Banach Space Theory: The Basis for Linear and Nonlinear Analysis'. I copy the question without altering anything. $\endgroup$ – Idonknow Dec 18 '13 at 18:38
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    $\begingroup$ 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. $\endgroup$ – Henning Makholm Dec 18 '13 at 18:39
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    $\begingroup$ 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. $\endgroup$ – TonyK Dec 18 '13 at 18:41
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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: $$(\alpha-\beta)\|y\| + \|\alpha x + \beta y\| = \alpha(\|x\|+\|y\|) .$$

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  • $\begingroup$ What motivates you to consider this approach? $\endgroup$ – Idonknow Dec 19 '13 at 5:37
  • $\begingroup$ It really was just symbol manipulation. Maybe I had seen this many years ago, and was regurgitating some long submerged memories. $\endgroup$ – Stephen Montgomery-Smith Dec 19 '13 at 5:40
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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.

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    $\begingroup$ 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 $\|\alpha x+\beta y\| = \alpha\|x\| + \beta\|y\|$. $\endgroup$ – Stephen Montgomery-Smith Dec 18 '13 at 23:56
  • $\begingroup$ Ah, that's a better interpretation of the question. $\endgroup$ – Theo Johnson-Freyd Dec 24 '13 at 5:07
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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 $\beta$ $\geq0$.

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  • $\begingroup$ 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. $\endgroup$ – Stephen Montgomery-Smith Dec 19 '13 at 0:35
  • $\begingroup$ 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. $\endgroup$ – Raul C. de Assis Dec 19 '13 at 0:41
  • $\begingroup$ OK. But how do you go from $\|\alpha x\| + \|\alpha y\| = |\alpha|\|x\| + |\beta|\|y\|$? $\endgroup$ – Stephen Montgomery-Smith Dec 19 '13 at 0:43
  • $\begingroup$ 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 $\mathbb{R}$. en.wikipedia.org/wiki/Norm_(mathematics) $\endgroup$ – Raul C. de Assis Dec 19 '13 at 0:45
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    $\begingroup$ Yes, but you changed an $\alpha$ on the LHS to a $\beta$ on the RHS. $\endgroup$ – Stephen Montgomery-Smith Dec 19 '13 at 0:47

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