Let $0< k_1\le k_2\dots \le k_m, k_i \in \mathbb{N}, k_1 \text{ an even positive integer }. f(x_1\dots x_m):=\sum_{i=1}^{m}{x_i}^{k_i}.$ I wanted to prove, if possible, that in a small enough ball $B(0;R),R<1$ we have:

$$|f(x_1\dots x_m)|\ge C||(x_1\dots x_m)||^{k_m}\forall x\in B(0,R)?$$

for some $C>0.$ What's $C?$

Sorry if it looks very easy, but how do we go about proving it, if true?

EDIT: It's starting to appear to me that the inequality may not be true after all, I'm writing down a counterexample below, could you please check and tell me if it works?

Take $m=2,$ so let's just work with $(x_1,x_2), f(x_1,x_2)=x_1^2 + x_2^3 (\text{therefore }k_1=2 { (even) }< k_2=3).$ Let's choose $(x_1,x_2)$ so that $f(x_1,x_2)=x_1^2 + x_2^3=0, e.g. x_1:=r^3, x_2=-r^2, r> 0.$ In this case $||(x_1,x_2)||=\sqrt{r^6 + r^4}.$ So clearly we can't have the inequality: $|f(x_1,x_2)|\ge C||(x_1,x_2)||^{k_2}, C>0.$

However, note that, here I really took advantage of the fact that either of $x_1,x_2$ can be negative. Let's now take a somewhat better example where things don't cancel: so take $x_1:=r^3, x_2:=-r^2 + r^3.$ In this case, a simple calculation shows that: ($x:=(x_1,x_2)$ below.)

$$\frac{x_1^2 + x_2^3}{||x||^3}=\frac{3r^7 -3r^8 +r^9}{r^6(1-2r +2r^2)^{3/2}}\to 0, \text{ as } r\to 0.$$

So again, the inequality can't be true here.

If all the $x_i's$ are indeed positive, then it seems to me that the inequality is indeed correct, here's a proof:

We know that $|a_1 + a_2|^p\le 2^{p-1}(|a|^p + |b|^p)$. This generalizes easily to: $|\sum_{i=1}^{m}a_i|^p\le 2^{m(p-1)}\sum_{i=1}^{m}|a_i|^p.$ Put $p:=k_m/2, a_i:=x_i^2$ to get: $||x||^p\le 2^{m(p-1)}\sum_{i=1}^{m}x_i^{k_m}\le 2^{m(p-1)}\sum_{i=1}^{m}x_i^{k_i}, x:=(x_1\dots x_m) \text{ near } 0, \text{ so } x_i^{k_i} \ge {x_i}^{k_m} \text{ as } k_i \le k_m \text{ and } x_i>0.$ The last line can be rearranged as $|f(x)|\ge \frac{1}{2^{n(p-1)}}||x||^{k_m}.$

Is the above correct? So the conclusion is that the inequality is true if all the components $x_i >0,$ but otherwise may not be true.

  • 1
    $\begingroup$ Your assumptions are not strong enough. If you have even one $k_i$ that is odd, then choosing $x_i=-R/2$ and $x_j=0$ for all $j \ne i$ will give you a function result that is negative, with postive $||x||$ and $ x \in B(0,R)$ $\endgroup$
    – Ingix
    Nov 29, 2023 at 18:13
  • $\begingroup$ @Ingix Thanks for bringing this to my attention - I completely forgot the modulus sing, which I just out back in. $\endgroup$ Nov 29, 2023 at 18:16
  • $\begingroup$ @Ingix I edited the question that now contains an answer in the negative and a partial answer in the affirmative. Could you please check? $\endgroup$ Nov 29, 2023 at 21:12

1 Answer 1


Your solution (both parts) looks good to me. I think the counterexample can be generalized straightforwardly if at least one $k_i$ is odd, so in this case there will always be a counterexample where $f(x)$ has value zero with $x=(x_1,\ldots,x_m)$ as close to, but not equal, $(0,\ldots,0)$ as desired.

If all $k_i$ are even, then your prove works again, as essentially you can replace each negative $x_i$ with $-x_i$ and the norm and $f$ doesn't change.

  • $\begingroup$ Thanks for checking my solution. Yes I also think that the counterexample will always work if at least one of the $k_i$ is odd. $\endgroup$ Nov 29, 2023 at 21:50

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