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?


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?

  • $\begingroup$ $x$ 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. $\endgroup$
    – Eoin
    Jun 7 '15 at 13:49
  • $\begingroup$ @Eoin, can you please read my question carefully once again? Your comment is not clear to me, I'm afraid. $\endgroup$ 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, $p(v)\geq 0$.

  • $\begingroup$ 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? $\endgroup$ Jun 7 '15 at 15:11
  • 1
    $\begingroup$ $v=\theta$ 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.... $\endgroup$ Jun 7 '15 at 15:30

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