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? 
 A: 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}
A: Note that
$|p(x)-p(y)|= p(x)-p(y)$ ou $p(y)-p(x)$, but as shown above we have
$$ 
p(x)-p(y)\leq p(x-y)\leq |p(x-y)|
$$
and
$$
p(y)-p(x)\leq p(y-x)\leq |p(y-x)|.
$$
On the other hand, by continuity of $p$ at $\theta$ we know that if $|| x-y||=||y-x||<\delta$, then 
$|p(x-y)|<\epsilon$ and too $|p(y-x)|<\epsilon$, and hence,
$$|p(x)-p(y)|<\epsilon,$$
which proves that $p$ is continuous at $x$.
A: 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.)
