Let $f$ be a continuous concave function on $[0,1]$ with $f(1)=0$ and $f(0)=1$. Does there exist a constant $k$ for which we can always draw a rectangle with area at least $k\cdot \int_0^1f(x)dx$, with sides parallel to the axes, in the area bounded by the two axes and the curve $f$?
If concavity is not required, it is possible to adapt from this example by using the curve $c/x$ to ensure that any rectangle has sufficiently small area. But with concavity, we know that $f$ lies above the line connecting the points $(0,1)$ and $(1,0)$, hence must have area at least $1/2$. If $f$ is exactly that line, then $k=1/2$ exactly. Otherwise, if $f$ is above the line, it looks like the rectangle will even get larger compared to the area under the curve.
Answer
Let $t\in[0,1]$ be a value such that $f(x)\le f(t)$ for all $x\in[0,1]$. (Such a value exists since $f$ is unimodal; I don't think we need continuity for this.) The triangle formed by $(0,0)$, $(1,0)$ and $(t,f(t))$ lies under the curve. Its area is $\frac12f(t)$, and it contains an axis-parallel rectangle with half its area, $\frac14f(t)$. The area under the curve is at most $f(t)$. Thus $k=\frac14$ suffices. I'm not sure this is the best possible constant, though.
No comments:
Post a Comment