Upper bound $a^k + (1-a)^k$

Question

For integer $k \geq 2$ and real $a \in [0, 1]$, I am looking for a function $f(k, a)$ such that

$$ a^k + (1-a)^k \leq f(k, a) $$

An easy one would be $f(k, a) = 2 \cdot (\max\{a, 1-a\})^k$. However, this bound is not good when $a$ is close to $1$ (or $0$).

Any thought would be helpful.

Side note: This problem came across while I was reading a paper.

Answer 1

For integer $k \ge 2$, you can get a quadratic bound just by forcing the parabola to match at 0, 1/2, and 1: $$ f(k,a) = 1-4(1-2^{1-k})a(1-a)\ge a^k +(1-a)^k $$ This is exact for $k = 2,3$ and an upper bound for $k > 3$ because a quadratic doesn't "bow out" as much as higher-order powers.

Answer 2

From Taylor around 0 and 1, $$ a^k + (1-a)^k \leq \min\{ 1-ka+k(k-1)a^2, 1-k(1-a)+k(k-1)(1-a)^2\} $$ but I don't know if this counts as 'simpler' for you

Answer 3

For $k\ge 2$, let $g_k$ be defined on $I=[0,1]$ by

$$ g_k(t)=t^k+(1-t)^k. $$

We have \begin{eqnarray} 0=g_k'(t)=k[t^{k-1}-(1-t)${k-1}] &\iff & t^{k-1}=(1-t)^{k-1}\cr & \iff& t=1-t \cr &\iff& t=0.5. \end{eqnarray}

Since $g''_k(0.5)=2^{1-k}>0$, $g_k$ has a minimum at $t=0.5$

It follows that the maximum of $g_k$ is achieved on the boundary of $[0,1]$. The maximum is then $$ g_k(0)=g_k(1)=1. $$

Answer 4

Erm... sorry, but for $a\in[0,1]$, the expression $a^k + (1-a)^k$ is decreasing as a function of $k$, so $a^2 + (1-a)^2$ would be the most obvious upper bound, since you want $k\ge2$.