# Non-negative non-decreasing polynomial lower than $x^N$ somewhere in $0 \le x \le 1$

Is there a degree $$N$$ polynomial $$f(x)$$ such that

• $$f(1) = 1$$,
• $$f(x)$$ is non-negative on $$0 \le x \le 1$$,
• $$f(x)$$ is non-decreasing on $$0 \le x \le 1$$, and
• there is a value of $$x$$ in $$0 \le x \le 1$$ such that $$f(x) < x^N$$?

My gut feeling is that there is not, but I don't know how to show that. (Edit: I don't have that gut feeling anymore after being shown it was wrong.)

For $$N\geq 3$$, there is such a polynomial. Take $$f(x)=x^{N-3}(x-3x^2+3x^3)$$. To show that this satisfies the desired property, it suffices to check the $$N=3$$ case.
• We compute $$f(x)=x-3x^2+3x^3=x(1-3x+3x^2)=x(x^3+(1-x)^3)\geq 0$$.
• We compute $$f'(x)=1-6x+9x^2=(3x-1)^2\geq 0$$.
• We see $$f(2/3)=2/9<8/27=(2/3)^3$$, so $$f(x) at $$x=2/3$$.
From here, the $$N\geq 3$$ case follows since multiplying by the nonnegative increasing polynomial $$x$$ does not disrupt any of the necessary properties.
For $$N=1$$, there is no such polynomial. The property that $$f(1)=1$$ implies that $$f(x)$$ would have to be $$1-a(1-x)$$ for some real $$a$$. If $$a>1$$ then $$f(0)<0$$, a contradiction, while if $$a\leq 1$$ then $$f(x)\leq x$$ for all $$x\in [0,1]$$.
For $$N=2$$, there is no such polynomial. Any quadratic $$f$$ besides $$x^2$$ can intersect the graph of $$x^2$$ at at most two points. One of these points is $$x=1$$; we know that $$f(0)\geq 0$$, so the other intersection point $$(x_0,x_0^2)$$ must satisfy $$x_0\in [0,1)$$. In particular, $$f(x) for $$x$$ slightly smaller than $$1$$, meaning that $$f'(1)\geq 2$$. Also, $$f'(0)\geq 0$$. Since $$f$$ is linear, this implies $$f(1)=f(0)+\int_0^1 f(t) dt=f(0)+\int_0^1 \big((1-t)f'(0)+tf'(1)\big)=f(0)+\frac{f'(0)+f'(1)}2.$$ This means $$1\geq f(0)+1$$, and so $$f(0)=0$$. Furthermore, as these inequalities are each tight, we must have $$f'(0)=0$$ and $$f'(1)=2$$; the only quadratic satisfying these conditions is $$f(x)=x^2$$.