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Let $X$ be a normed linear space, $M=\{x:x=\alpha\,x_0,\;\alpha\;\text{is a scalar}\},\;x_0\neq 0$ be a member of $X$ and define \begin{align} f:&M\to \Bbb{R},\\& x \mapsto f(x)=\alpha\|x_0\|.\end{align} I want to prove that $f$ is a linear functional on $M$.

HERE'S WHAT I'VE DONE

Let $\gamma,\,\eta\in \Bbb{R}$ and $x,y\in M,$ then, \begin{align} f(\gamma x+\eta y)&=\alpha\|\gamma x_0+\eta y_0\|\\&\leq \alpha\left(\|\gamma x_0\|+\|\eta y_0\|\right)\\&= \alpha|\gamma |\|x_0\|+\alpha |\eta |\| y_0\|\\&= |\gamma |f(x)+|\eta |f(y).\end{align} I'm stuck here but I think I'm missing something. Can you please share with me, what I'm missing?

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    $\begingroup$ The $y_0$ should be an $x_0$. $\endgroup$ – Kenny Wong Dec 29 '18 at 9:53
  • $\begingroup$ @Kenny Wong: Really? $\endgroup$ – Omojola Micheal Dec 29 '18 at 9:53
  • $\begingroup$ @Mike what is $y_0$ $\endgroup$ – mathworker21 Dec 29 '18 at 9:54
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    $\begingroup$ You're meant to say, if $x = \alpha x_0$, and $y = \beta x_0$ (for some $\alpha, \beta \in \mathbb R$), then $f(\gamma x + \eta y) = f ((\gamma \alpha + \eta \beta) x_0) = (\gamma \alpha + \eta \beta) \| x_0 \|$... $\endgroup$ – Kenny Wong Dec 29 '18 at 9:55
  • $\begingroup$ @Kenny Wong: Thanks a lot! $\endgroup$ – Omojola Micheal Dec 29 '18 at 9:56
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Credits to Kenny Wong:

Let $\gamma ,\eta \in \Bbb{R}$ and $x,y\in M,$ then, $x=\alpha x_0$ and $y=\beta x_0,$ for some $\alpha,\beta\in \Bbb{R}.$ So, \begin{align} f(\gamma x+\eta y)&=f(\gamma (\alpha x_0)+\eta (\beta x_0))\\&=f((\gamma \alpha +\eta \beta) x_0)\\&=(\gamma \alpha +\eta \beta) \|x_0\|\\&=\gamma \alpha \|x_0\| +\eta \beta \|x_0\|\\&=\gamma f( x)+\eta f( y).\end{align} Since $x,y\in M$ and $\gamma ,\eta \in \Bbb{R}$ are arbitrary, then \begin{align} f(\gamma x+\eta y)=\gamma f( x)+\eta f( y),\;\;\forall\;x,y\in M,\;\forall\; \gamma ,\eta \in \Bbb{R}.\end{align} Thus, $f$ is a linear functional on $M$.

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