$\max_{\|x\|=1}f(x)=\max_{\|x\|\le 1} f(x)$ I'm reading the book "Matrix Analysis and Applied Linear Algebra". On page 450, eq(5.15.5), I think I found an error made by the author. So I post it here. If I'm wrong, please correct me.
The statement from the book: If $f: \mathbb{R}^n\rightarrow\mathbb{R}$ satisfies $f(ax)=af(x)$ for any $a\ge0$, then
$$\max_{\|x\|=1}f(x)=\max_{\|x\|\le 1} f(x)$$
I believe the above statement I get from the book is wrong. A conterexample is $f(x)=-\|x\|$.
In order to make the statement correct, we can to add the condition: $f(x)\ge0$. If $f(x)\ge0$, then it is straightforward to prove.
Now I think I have a more interesting question. I cannot figure out if the following similar statement is correct.
A new and similar statement: If $f: \mathbb{R}^n\rightarrow\mathbb{R}$ satisfies $f(ax)=af(x)$ for any $a\ge0$, and $\max_{\|x\|=1}f(x)\ge0$, then
$$\max_{\|x\|=1}f(x)=\max_{\|x\|\le 1} f(x)$$
Can any one tell me whether this statement is correct? Thanks.
 A: Yes, the new statement is correct.  If $f(x) \ge 0$ and $\|x\| \le 1$, then $$f\left(\frac{x}{\|x\|}\right) = \frac{f(x)}{\|x\|} \ge f(x)$$
A: Yes, it's correct. Observe that $f(\mathbf{0})=f(0\cdot\mathbf{0})=0f(\mathbf{0})=0$. There are two possibilities:


*

*If $f(\mathbf{0})=\max_{\|x\|\le 1} f(x)$, we have $0=f(\mathbf{0})=\max_{\|x\|\le 1} f(x)\ge\max_{\|x\|=1}f(x)\ge0$ and the result follows.

*If $f(u)=\max_{\|x\|\le 1} f(x)$ for some $u\neq\mathbf{0}$, we must have $f(u)\ge f(\mathbf{0})=0$. Hence $f(\frac{u}{\|u\|}) = \frac{f(u)}{\|u\|}\ge f(u)=\max_{\|x\|\le 1} f(x)$ and the result follows.

A: user1551 has given the correct answer. Here I just state it in another way.
Proposition 1: if $f(ax)=af(x)$ for any $a\ge0$, then
$$\max_{\|x\|\le1}f(x)=\max_{\|x\|=1}f(x) \text{ or } \max_{\|x\|\le1}f(x)=f(0)$$
Proof: If $f(0)\ge f(x)$ for all $\|x\|\le1$, clearly $\max_{\|x\|\le1}f(x)=f(0)$.
Next suppose there exists a nonzero vector $x_0$ such that $\|x_0\|\le1$ and
$$f(x_0)\ge f(x)$$
for any other $x$ with $\|x\|\le1$.
Then as a special case, we have
$$f(x_0)\ge f(x_0/\|x_0\|)=f(x_0)/\|x_0\|$$
Note $f(x_0)\ge f(0)=0$. 


*

*(i) if $f(x_0)>0$, then the above inequality implies that $\|x_0\|=1$
and hence $\max_{\|x\|\le1}f(x)=\max_{\|x\|=1}f(x)$; 

*(ii) if $f(x_0)=0$, then $\max_{\|x\|\le1}f(x)=f(0)$.


Q.E.D.
Corollary 1: If $f(ax)=af(x)$ and $\max_{\|x\|=1}f(x)\ge0$, then $\max_{\|x\|=1}f(x)=\max_{\|x\|\ge1}f(x)$.
Corollary 2: If $f(ax)=af(x)$ and $\max_{\|x\|\le1}f(x)>0$, denote $x_0=\arg\max_{\|x\|\le1}f(x)$, then $\|x_0\|=1$.
Remark: Proposition 1 does not imply that the maximum point is either $x=0$ or $\|x\|=1$. It is still possible that the maximum point $x_0$ satisfies $0<\|x_0\|<1$. But in that case we must have $f(x_0)=0$.
