# Continuity of Minkowski functional in locally convex topological vector space

Let $X$ be a locally convex topological vector space over $\mathbb{R}$ or $\mathbb{C}$ and let $p_C(x)=\inf (\lbrace t>0 \mid t^{-1}x \in C\rbrace)$ be the Minkowski functional for an arbitrary open convex neighbourhood $C$ of $0$ in $X$. I need to show that for any $C$ as above that the Minkowski functional is continuous, but I have a hard time getting started on the proof, since I can't figure out what the pre-image of the Minkowski functional is for an arbitrary open subset of the positive real line $[0,\infty)$. Can I show it without assuming that $C$ is balanced?

• you can see W. Rudin, Functional analysis . P26 – Ali Mar 15 '16 at 13:26
• Thank you, I guess the gist of the argument is that $p_C$ is bounded by $1$ in $C$ so therefore continuous. – Bartuc Mar 16 '16 at 9:02
• Can you elaborate on what exactly I need. The theorem you refer to (1.34 in Rudin), doesn't seem to tell that the Minkowsky functional is continuous, but it shows more that it's a semi-norm, but is there a way around that? I.e can I show it more directly, without assuming or proving that it is a semi-norm? – Bartuc Mar 18 '16 at 12:18
• The following property of seminorms: $$p(x)-p(y)\leq |p(x-y)|$$ showa that $p$ is continuous. – Ali Mar 18 '16 at 12:30
• Yes, but what if I want to show it without showing that Minkowski is a seminorm? I have not assumed that $C$ is balanced (which is equivalent to $p_C$ being a semi-norm). I there really no direct way, without going through seminorms? I just don't think I can take for granted that $X$ is a semi-normed space, without loss of generality. – Bartuc Mar 18 '16 at 12:49

It can be easily shown that every Minkowski functional functional satisfy in$$p_c(x-y)\leq p_c(x)-p_c(y)$$

• As far as I know, the Minkowski functional satisfies the following properties: (1) $p_C(x+y) \leq p(x)+p(y)$ (2) $p_C(tx)=tp_C(x) \, , \, \forall t \geq 0$ (3) $x \in C \Leftrightarrow p_C(x)<1$ From (1) I can prove that $p_C(x)-p_C(y) \leq p_C(x-y)$ via the inequalities $p_C(x)=p_C(x-y+y) \leq p_C(x-y)+p_C(y)$ and subtracting $p_C(y)$ on both sides. How do you get the reverse inequality? – Bartuc Mar 19 '16 at 12:51
• You Can deduce that from this. Use $y-x$ instate of $x-y$ you can deduce that $p(x)-p(y)\leq p(x-y).$ – Ali Mar 19 '16 at 12:55
• Okay I can see that. Rudin writes the inequality as $\vert p(x)-p(y) \vert \leq p(x-y)$ if $p$ is a semi-norm and on page 27 he gives a proof of their continuity. What if I insist on proving it directly via the definition of continuity? I just don't want to prove it by first proving it's a seminorm, surely there must some way to do so? – Bartuc Mar 19 '16 at 13:07

Here is a try for how to solve it.

Since the topology on $\mathbb{R}$ is generated by the open intervals $(a,b)$ with $a<b$, the subspace topology on $[0, \infty)$ is generated by the sets of the form $(a,b)\cap [0, \infty)$ it suffices to prove that $p_C^{-1}((a,b)\cap [0, \infty))$ is open in $X$. Now, since the topology on $X$ is generated by sets of the form $p+C$ for $C \in \gamma$ and $p \in X$, where $\gamma$ is local base of convex sets $C$, it is enough to show that for any $x \in p_C^{-1}((a,b)\cap [0, \infty))$, there exists a $C' \in \gamma$ such that $x + C' \subseteq p_C^{-1}((a,b)\cap [0, \infty))$.

Now, there are four possibilities for what the set $(a,b)\cap [0, \infty)$ is:

$(a,b)\cap [0, \infty)=(a,b)$ if $b>a>0$

$(a,b)\cap [0, \infty)=(0,b)$ if $a=0$

$(a,b)\cap [0, \infty)=[0,b)$ if $a<0$ and $b>0$

$(a,b)\cap [0, \infty)=\emptyset$ if $b\leq0$

Take for example the first case $p_C^{-1}((a,b))$:

Let $x_0 \in p_C^{-1}((a,b))$, then $a<p_C(x_0)<b$, since $x_0+C$ is an open neighborhood of $x_0$ from the base for the topology on $X$, we have the required open set. Now here's the tricky part: I want to show that $x_0+C \subseteq p_C^{-1}((a,b))$, but I run into trouble because of the following analysis ($c_0 \in C$):

$p_C(x_0+c_0)\leq p_C(x_0)+p_C(c_0) < b+1$

But since an element in $p_C^{-1}((a,b))$ has to satisfy "$<b$", this does not work in this case (or the others?). Any help would be much appreciated.