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If $p$ is a sublinear functional on a real vector space $X$, then there exists a linear functional $\tilde{f}$ on $X$ such that $-p(-x) \leq \tilde{f}(x) \leq p(x)$ for all $x \in X$.

How to prove this result?

For all $x, y \in X$, we have $p(x+y) \leq p(x) + p(y)$.

And, for all $x \in X$ and for all $\alpha \in \mathbb{R}$ such that $\alpha > 0$, we have $p(\alpha x) = \alpha p(x)$.

These two conditions imply that $p(\theta) = 0$, where $\theta$ denotes the zero vector in $X$, and $-p(-x) \leq p(x)$ for all $x \in X$.

What next?

How to proceed from here?

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Take any non-zero $x\in X$ and let $Y=Span\{x\}$. Now define a linear functional $f$ on $Y$ as follows \begin{equation} f(\alpha x) =\alpha p(x). \end{equation} If $\alpha >0$ then \begin{equation}f(\alpha x) = \alpha p(x) = p(\alpha x),\end{equation}and if $\alpha <0$, then \begin{align}f(\alpha x) &= \alpha p(x)\\ & \leq -\alpha p(-x) & & (\text{because $-p(-x)\leq p(x)$ and $\alpha<0$})\\ & = p(\alpha x). \end{align} Therefore, $f(y) \leq p(y)$ for all $y\in Y$. Now by Hahn-Banach Theorem you can extend $f$ to a linear functional $\tilde{f}$ on $X$ such that $\tilde{f}(x) \leq p(x)$ for all $x\in X$.

Moreover, by linearity of $\tilde{f}$ you get that $-\tilde{f}(x) = \tilde{f}(-x) \leq p(-x) \Rightarrow \tilde{f}(x) \geq -p(-x)$. Combining these things together gives $-p(-x) \leq \tilde{f}(x) \leq p(x)$ for all $x\in X$.

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  • $\begingroup$ thanks. So nice of you! If you don't mind my asking this, are you answering my questions by consulting some anciliary material? Or, are you answering these questions just off-hand? $\endgroup$ – Saaqib Mahmood Jun 7 '15 at 18:43
  • $\begingroup$ I solved these questions when I was learning functional analysis from Kreyszig....and your questions help me to revise the material.... $\endgroup$ – Urban PENDU Jun 7 '15 at 19:49
  • $\begingroup$ wonderful. Did you solve them on your own? Or, did you have some tutorial support? How much of Kreyszig's book have you studied in class? And how much of it have you learnt by yourself? $\endgroup$ – Saaqib Mahmood Jun 7 '15 at 20:32
  • $\begingroup$ kreyszig was not followed in the class...I did it by myself. I like kreyszig's presentation of the subject. In class we followed Conway's book on functional analysis $\endgroup$ – Urban PENDU Jun 7 '15 at 20:50

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