# 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?

• $x$ 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. – Saaqib Mahmood 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, $$0\leq p(nv) = p(v+\ldots+v)\leq np(v).$$ 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? – Saaqib Mahmood 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.... – Urban PENDU Jun 7 '15 at 15:30