# Prob. 10, Sec. 4.2 in Kreyszig's functional analysis book: There is a linear functional for every sublinear functional …

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?

Take any non-zero $x\in X$ and let $Y=Span\{x\}$. Now define a linear functional $f$ on $Y$ as follows $$f(\alpha x) =\alpha p(x).$$ If $\alpha >0$ then $$f(\alpha x) = \alpha p(x) = p(\alpha x),$$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 $x\in X$.