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I know that the graph of continuous functions on $[0,1]$ is a null set (has Lebesgue measure zero). My question is following and seems relate to that I have conceded. Let $f:[0,1]\to[0,1]$ be a continuous and strictly increasing function. Then for any $n\in\mathbb N$, we can cover the graph of $f$ by $n$ rectangle, with side parallel x-and y-axes, such that each rectangle has area $=\frac1{n^2}$. (Rectangles can be overlap).

How to prove this? Any idea?

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    $\begingroup$ Unclear what you are asking. Please re-write carefully so someone can start thinking of an answer. $\endgroup$ – Mathemagical Dec 16 '17 at 14:16
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First, to prevent the notation from exploding, extend $f$ to an increasing function $f:[0,\infty)\to[0,\infty)$ which tends to infinity at infinity.

Since $f$ is increasing,

Note: If $0\le a<b$ then the rectangle $[a,b]\times[f(a),f(b)]$ covers the part of the graph of $f$ lying in the strip $a\le x\le b$.

Now let $x_0=0$. Since $f$ is increasing and continuous we can recursively choose $x_1,x_2,\dots$ so that $x_{j+1}>x_j$ and $$(x_{j+1}-x_j)(f(x_{j+1})-f(x_j))=\frac1{n^2}.$$

If we can show $x_n\ge1$ we're done; then we've covered the graph of $f$ by $n$ rectangles of area $1/n^2$. Since $f([0,1])\subset[0,1]$ it's enough to show that $x_n\ge1$ or $f(x_n)\ge1$ (because if $x_n<1$ then also $f(x_n)< 1$, which is to say $f(x_n)\ge1$ implies $x_n\ge1$).

But Cauchy-Schwarz shows that $$1=\sum_{j=0}^{n-1}(x_{j+1}-x_j)^{1/2}(f(x_{j+1}-f(x_j))^{1/2}\le\left(\sum(x_{j+1}-x_j)\sum(f(x_{j+1}-f(x_j))\right)^{1/2}=\left((x_n)(f(x_n)-f(0)\right)^{1/2};$$hence either $x_n\ge1$ or $f(x_n)-f(0)\ge 1$.

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