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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.

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  • $\begingroup$ $p$ doesn't have to be a norm, since it could be identically zero. $\endgroup$ – carmichael561 Mar 21 '18 at 21:23
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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''.

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Hint: the property iv) enables you to show that $p$ is continuous: just consider the inverse image $U$ of an open interval and suppose it is not open, take $x= lim_nx_n, x_n \in U$, and $x$ is not in $U$, deduce the contradiction.

Deduce that $p$ is bounded in a ball and use ii) to show the existence of M.

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  • $\begingroup$ Thanks! Do you mean to take $x_n \notin U$ and $x \in U$, however? This would derive the contradiction in the assumption that the complement of $U$ is not closed. $\endgroup$ – Greg K Mar 21 '18 at 21:52

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