Prob. 6, Sec. 4.2 in Kreyszig's functional analysis book: Continuity of a subadditive functional at zero implies continuity

Let $X$ be a real normed space, and let $p \colon X \to \mathbb{R}$ be a functional such that $$p(x+y) \leq p(x) + p(y) \ \mbox{ for all } \ x, y \in X$$ and such that $$p(\theta) = 0, \ \mbox{ where \theta denotes the zero vector in X.}$$ Suppose that $p$ is continuous at $\theta$.

Then how to show that $p$ is continuous at any point $v$ of $X$?

Since $p$ is continuous at $\theta$, given $\epsilon > 0$, there is a $\delta >0$ such that, for all $x \in X$ with $\Vert x - \theta \Vert < \delta$, we have $$\vert p(x) - p(\theta) \vert = \vert p(x) \vert < \epsilon.$$

What next?

First note that

The subadditivity of $p$ shows that $$p(x) = p(x - y + y) \leq p(x-y) + p(y)$$ so that $p(x)-p(y) \leq p(x-y)$. This also shows that $p(y) - p(x) \leq p(y-x)$. Hence $$-p(y-x)\leq p(x) - p(y) \leq p(x - y).$$

Let $x\in X$ be any point. Showing continuity of $p$ at $x$ means that given any $\epsilon>0$ there exist $\delta >0$ such that $|p(x)-p(y)|<\epsilon$ whenever $\|x-y\|<\delta$. So, let $\epsilon>0$. Then by continuity of $p$ at $\theta$ we know that there exist $\delta>0$ such that $|p(z)|<\epsilon$ whenever $\|z\|<\delta$. Now, whenever $\|x-y\|<\delta$ then note that $p(x-y)<\epsilon$ and similarly $-\epsilon <-p(y-x)$ and hence \begin{align} -\epsilon <-p(y-x)\leq p(x)-p(y)\leq p(x-y) <\epsilon & & \Rightarrow & & |p(x)-p(y)|<\epsilon. \end{align}

• Hello, @Urban PENDU? Thank you for your answer. But how do you conclude that $p(x-y) = p(y-x)$? – Saaqib Mahmood Jun 7 '15 at 9:35
• regarding the question in my last comment. Since $0 = p(\theta) = p(x-x) = p(\ x + (-x) \ ) \leq p(x) + p(-x)$, we have $-p(x) \leq p(-x)$; so $-p(-x) \leq p(\ -(-x) \ ) = p(x)$. What next? – Saaqib Mahmood Jun 7 '15 at 9:46
• sorry i took $p$ to be a seminorm....I will edit the argument.... – Urban PENDU Jun 7 '15 at 9:47
• please do. Are you doing an MPhil in mathematics? Which phase are you in ---- research or coursework? Are you from Punjab in India? – Saaqib Mahmood Jun 7 '15 at 9:51
• made an edit...please have a look and if there is any mistake let me know...I think the argument is fine now....I am doing M.Sc Mathematics – Urban PENDU Jun 7 '15 at 9:55

First, a hint. Try considering the difference $p(x)-p(y)$. Can you show this gets small as $x\rightarrow y$? (Remember the standard trick of adding $0$ in a suitable form.)

• That's just what I'm struggling to achieve, @Stromael. So can you please elabotate on your hint? – Saaqib Mahmood Jun 7 '15 at 9:11