Let $(X,\lVert \cdot \rVert)$ be a (real) Banach space and let $p$ be a (real) functional on $X$ satisfying the following conditions:
(i) $p(x) \geq 0$ for all $x \in X$
(ii) $p(\alpha x) = \alpha p(x)$ for all $x \in X$ and $\alpha > 0$
(iii) $p(x_1 + x_2) \leq p(x_1) + p(x_2)$ for all $x_1, x_2 \in X$
(iv) if $x_n \rightarrow x$ in $X$, then $p(x) \leq \lim p(x_n)$
Show that there is an $M > 0$ so that $p(x) \leq M \lVert x \rVert$ for all $x \in X$.
I think this is going to come down to showing that $p$ is a norm, that $X$ is complete with respect to the metric defined by that norm, and then applying the open mapping theorem to the identity map $(X,\lVert \cdot \rVert) \rightarrow (X,p)$, but I'm not sure how to show that $p$ gives a norm. I can show that $p(0) = 0$ (consider the sequence $\frac{1}{n} x$ for some nonzero $x$) and the triangle inequality follows straight from the properties of $p$, but I'm not sure how to show that $p(-x) = p(x)$ or how to show that $p(x) = 0$ implies $x = 0$.
Note: this isn't a homework problem, it's an old qual problem that I'm using to study for an exam.
If property (iii) means that for all converging sequences $x_n\to x$ the limit $p(x_n)$ exists and satisfies $p(x) \le \lim p(x_n)$ then the claim can be proved by a simple contradiction argument:
Assume the claim is not true. Then for each $n$ there is $x_n$ such that $p(x_n) > n^2\|x_n\|$. Since $p(0)=0$ by (ii)(iv) it follows $x_n\ne0$. Due to (ii) we can divide the inequality by $\|x_n\|$. Thus, wlog $\|x_n\|=1$.
Dividing the inequality by $n$ yields $$ p(n^{-1} x_n) > n. $$ Now $n^{-1}x_n\to0$, so that $\lim p(n^{-1}x_n)$, exists, which is a contradiction to the inequality, where we got that $p(n^{-1}x_n)$ is unbounded.
Note that you can replace (iii) by ``for all $x_n\to x$ the limit $p(x_n)$ exists''.