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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?

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First note that

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

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}

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  • $\begingroup$ Hello, @Urban PENDU? Thank you for your answer. But how do you conclude that $p(x-y) = p(y-x)$? $\endgroup$ – Saaqib Mahmood Jun 7 '15 at 9:35
  • $\begingroup$ 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? $\endgroup$ – Saaqib Mahmood Jun 7 '15 at 9:46
  • $\begingroup$ sorry i took $p$ to be a seminorm....I will edit the argument.... $\endgroup$ – Urban PENDU Jun 7 '15 at 9:47
  • $\begingroup$ please do. Are you doing an MPhil in mathematics? Which phase are you in ---- research or coursework? Are you from Punjab in India? $\endgroup$ – Saaqib Mahmood Jun 7 '15 at 9:51
  • $\begingroup$ 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 $\endgroup$ – Urban PENDU Jun 7 '15 at 9:55
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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.)

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

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