Prob. 8, Sec. 4.2 in Kreyszig's functional analysis book: Nonnegativity of a subadditive functional outside a sphere implies nonnegativity

If a subadditive functional $$p$$ defined on a normed space $$X$$ is non-negative outside a sphere $$\{ \ x \in X \ \colon \ \Vert x \Vert = r \ \}$$, then how to show that $$p$$ is non-negative for all $$x \in X$$?

For all $$x, y \in X$$, we have $$p(x+y) \leq p(x) + p(y)$$.

And, $$p(x) \geq 0$$ for all $$x \in X$$ such that $$\Vert x \Vert > r$$, where $$r$$ is a given positive real number.

Let $$v \in X$$ be arbitrary. Let $$v$$ be non-zero.

Let $$x \colon = \frac{r+1}{\Vert v \Vert } v.$$ Then $$\Vert x \Vert = r+1$$. So we must have $$p(x) \geq 0$$.

What next?

PS:

Here is a solution:

First, we show that $$n p(x) \geq p(nx) \ \mbox{ for all } n \in \mathbb{N}. \tag{0}$$ For $$n = 1$$, this holds trivially. So if this holds for any given $$n \in \mathbb{N}$$, then we find that $$(n+1)p(x) = np(x) + p(x) \geq p(nx) + p(x) \geq p(nx+x) = p\big( (n+1) x \big).$$ Hence (0) holds for all natural numbers $$n$$.

We are given that, there is a (non-negative) real number $$r$$ such that $$p(x) \geq 0 \ \mbox{ for all } x \in X \mbox{ such that } \lVert x \rVert > r. \tag{1}$$

Let $$x \in X$$ such that $$\lVert x \rVert \leq r$$. There are two cases according as $$\lVert x \rVert > 0$$ or $$\lVert x \rVert = 0$$.

Case 1: If $$\lVert x \rVert > 0$$, then let us choose a natural number $$n$$ such that $$\lVert nx \rVert = n \lVert x \rVert > r.$$ This along with (0) amd (1) above yields $$n p(x) \geq p(nx) \geq 0,$$ and as $$n > 0$$, so we obtain $$p(x) \geq 0.$$

Before considering the case $$\lVert x \rVert = 0$$, we note that, for any $$x, y \in X$$, we have $$p(x) = p(x - y + y) \leq p(x-y) + p(y),$$ and so $$p(x) - p(y) \leq p(x-y),$$ which is the same as $$p(x-y) \geq p(x) - p(y). \tag{2}$$

Case 2: If $$\lVert x \rVert = 0$$, then $$x = \mathbf{0}_X$$, the zero vector in $$X$$. In this case we use (2) above and find that, for any $$x \in X$$, $$p\left( \mathbf{0}_X \right) = p\big(x - x \big) \geq p(x) - p(x) = 0,$$ and so $$p\left( \mathbf{0}_X \right) \geq 0,$$ as required.

Is this proof correct? If so, is it clear enough?

• $x$ nonnegative outside of the sphere. Try to write $p(x)<\sum p(1/a x)$ to see that everything in the sphere is nonnegative as well.
– Eoin
Jun 7 '15 at 13:49
• @Eoin, can you please read my question carefully once again? Your comment is not clear to me, I'm afraid. Jun 7 '15 at 14:17

You are almost there...you might have just missed a simple trick. Note that in your solution you have not yet used the fact that $p$ is a subadditive functional. Here is how you can use it. For any $v\in X$ let $n$ be an integer such that $n\geq \frac{r+1}{\|v\|}$. Then it follows that $p(nv)\geq 0$. In particular, \begin{equation} 0\leq p(nv) = p(v+\ldots+v)\leq np(v). \end{equation} Therefore, $p(v)\geq 0$.
• and how to tackle the case when $v = \theta$, the zero vector in $X$? Did they make any vidoes of your B. Maths (Hons.) courses? Jun 7 '15 at 15:11
• $v=\theta$ is not a problem because by taking $x=y=\theta$ you can easily see that $p(\theta)\geq0$....and there were no videos made of the class lectures.... Jun 7 '15 at 15:30