Continuous function with $f(x^m+y^n) \le f(x+y) $

Question

Let $m>n\ge 1$ be integers. If $m$ is even and $f:\mathbb{R} \to \mathbb{R} $ is continuous, nonconstant, with $f(x^m+y^n) \le f(x+y) $, $\forall x, y \in \mathbb{R} $, prove that $n$ is even.
I think that we should suppose $n$ is odd and reach some kind of contradiction. By taking $x=0$ we get that $f(y) \ge f(y^n), \forall y\in \mathbb{R} $, but I don't know how to use this.
EDIT: Is there any chance that the problem is wrong? An user in the comments pointed out that the fact that $m$ is even is redundant. This makes me doubt a little the correctness of the problem, but I can't make any progress in this direction either.
Bump

Answer 1

Not a solution of the problem - but, at the request of the OP, an observation showing that the assumption that $m$ is even follows from other assumptions (which suggests that the problem may be incorrectly stated).

My claim is:

If $m\ge n>1$ are integers, and $f\colon\mathbb R\to\mathbb R$ is non-constant and satisfies $f(x^m+y^n)\le f(x+y)$ for any $x,y\in\mathbb R$, then $m$ is even.

(Notice that continuity is not required.)

For the proof, it suffices to show that if $m$ is odd, then for any $a,b\in\mathbb R$ there exist $x,y\in\mathbb R$ such that $x^m+y^n=a$ and $x+y=b$. Indeed, from the second equation $y=b-x$, and substituting this into the first equation we get $x^m+(b-x)^n=a$. The LHS is a polynomial in $x$ of degree $m$; since $m$ is assumed to be odd, the range of values of this polynomial is the whole $\mathbb R$. This means that the equation $x^m+(b-x)^n=a$ has a solution. Substituting this solution into $y=b-x$, we recover $y$.

Answer 2

As per request, I'm posting a partial effort. Throughout (despite my comment), $m$ is assumed even.

Fix some $y \in (-1, 1)$, and define $a_k = y^{n^k}$. Note that $a_k \to 0$ as $k \to \infty$. Note also that, from the $x = 0$ case, we see that $$f(a_{k+1}) = f(a_k^n) \le f(a_k).$$ Thus, using continuity of $f$, $$f(y) = f(a_0) \le \lim_{k\to \infty}f(a_k) = f\left(\lim_{k\to \infty}a_k\right) = f(0),$$ That is, $f$ restricted to the interval $(-1, 1)$ achieves its maximum at $0$. Note that, by continuity, this interval can be extended to $[-1, 1]$.

Now, consider the inequality when $x = 1$. We get $$f(y^n + 1) \le f(y + 1).$$ Fix $y \in (0, 2)$, and let $b_k = (y - 1)^{n^k} + 1$. Then $b_k \to 1$, and $$f(b_{k+1}) = f((b_0 - 1)^{n^k})^n + 1) \le f((b_0 - 1)^{n^k} + 1) = f(b_k).$$ As with the previous argument, this implies that the maximum of $f$, restricted to $[0, 2]$, is achieved at $1$.

Combining this with the previous conclusion, this means that $f(0) = f(1)$, as they are both maxima over the interval $[0, 1]$. Further, taking the original inequality with $x = -1$ and $y = 0$, we get $f(0) = f(1) \le f(-1)$, but as previously discussed, $0$ maximises $f$ over $[-1, 1]$, hence $f(-1) = f(0) = f(1)$.

If we restrict our attention to $y \in (0, 1)$, we can extend the sequence $a_k$ as above for negative integers $k$ (as $y$ is positive, we need not yet worry about $n$ being odd). We still have $f(a_{k + 1}) \le f(a_k)$. Note that $\lim_{k \to -\infty} a_k = 1$, so $$f(y) = f(a_0) \ge \lim_{k \to -\infty} f(a_k) = f\left(\lim_{k \to -\infty} a_k\right) = f(1).$$ This implies $f$ is constant over $[0, 1]$ (and all of this without assuming anything extra about $n$).

Finally, suppose $n$ is odd. This means we can consider $a_k$ for negative $k$, even when $y \in (-1, 0)$. This time we have $\lim_{k \to -\infty} a_k = -1$, hence as above, $f(y) \ge f(-1) = f(0)$, which implies $f$ is constant over $[-1, 1]$.