Is this following inequality with increasing powers of the components true for small $x$? And if yes, what's the positive constant $C?$

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.

• 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 $x \in B(0,R)$ Nov 29, 2023 at 18:13
• @Ingix Thanks for bringing this to my attention - I completely forgot the modulus sing, which I just out back in. Nov 29, 2023 at 18:16
• @Ingix I edited the question that now contains an answer in the negative and a partial answer in the affirmative. Could you please check? Nov 29, 2023 at 21:12

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.
• 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. Nov 29, 2023 at 21:50