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Let $h,k:[a,b]\to \mathbb R$ be two $C^1$ functions which coincides at the extremes and such that $h\leq k$ on $[a,b]$ and let the graph of $h$ be an arc of a circle of radius $R$ curved upwards. Show that $$L(k)-L(h)\geq \frac 1R|D|,$$ Where $L(k)$ and $L(h)$ denotes the lengths of the graph of $k$ and $h$ respectively, and $|D|$ is the area of the region $D$ between the two graphs.

My attempt: Playing around and recalling that the length of a graph on an interval is related to the quantity $$\sqrt{1+[f'(x)]^2},$$ I've studied the following inequality $$\sqrt{1+y^2}-\sqrt{1+x^2},$$ and it seems to me that, using the mean value theorem, and the fact that $$\left(\frac{x}{\sqrt{1+x^2}}\right)^'=\left(\frac{1}{(1+x^2)^{3/2}}\right)>0$$ that the following inequality holds, $\forall x,y\in\mathbb R$:

$$\sqrt{1+y^2}-\sqrt{1+x^2}\geq \frac{x}{\sqrt{1+x^2}}(y-x).$$ Please correct me if I am wrong. My plan now is to use this inequality to recover information about the inequality of the problem, but i cannot go further.

Any help is appreciated. Thank you very much.

Edit: I admit your objection is valid, thanks @Beni, i add the hypothesis. And now? How to prove the statement which now should be well posed I think.

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Surely, you must have left out a condition, namely that the graph of $h$ must be an arc of the upper half of a circle. Otherwise, the left side of the inequality can be negative. –  Harald Hanche-Olsen Mar 11 '12 at 13:40
    
The claim is false. E.g. take the graph of $k$ to be a straight line, so that $L(k)<L(h)$. –  Chris Eagle Mar 11 '12 at 13:42
    
@Both I think your objections are not possible since it is written that $h\leq k$ on $[a,b]$. I'm I right? –  uforoboa Mar 11 '12 at 14:02
    
@uforoboa: If $h$ is not curved upwards, then the length of $k$ may be smaller than the length of $h$. Make a few drawings. The inequality you intend to use is correct. Not sure how that will help you, but is a good start. –  Beni Bogosel Mar 11 '12 at 14:07
    
@All.. Thanks for your attention. I've modified the statement according to your observations –  uforoboa Mar 11 '12 at 14:34
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2 Answers

up vote 2 down vote accepted

Let $h,\ k:[a,b]\to \mathbb{R}$ be two $C^1(a,b)$ functions s.t.:

  1. $k\geq h$ in $[a,b]$, $k(a)=h(a)$ and $k(b)=h(b)$;
  2. the graph of $h$ is a concave arc of a circle, that is there esist $R>0$, $x_0\in [a,b]$ and $y_0\in \mathbb{R}$ s.t.: $$\tag{0} h(x) = y_0 +\sqrt{R^2 - (x-x_0)^2}\; .$$

Now, as in the OP, from the convexity inequality: $$\forall t,T\in \mathbb{R},\quad \sqrt{1+T^2}\geq \frac{t}{\sqrt{1+t^2}}\ (T-t) +\sqrt{1+t^2}$$ we get: $$\tag{1} \mathcal{L}(k) - \mathcal{L}(h) \geq \int_a^b \frac{h^\prime}{\sqrt{1+(h^\prime)^2}}\ (k^\prime -h^\prime)\; .$$ An integration by parts of the rightmost side of (1) yields: $$\tag{2} \begin{split} \int_a^b \frac{h^\prime}{\sqrt{1+(h^\prime)^2}}\ (k^\prime -h^\prime) &= \frac{h^\prime}{\sqrt{1+(h^\prime)^2}}\ (k-h) \Bigg|_a^b - \int_a^b \left( \frac{h^\prime}{\sqrt{1+(h^\prime)^2}} \right)^\prime\ (k-h) \\ &= \frac{1}{R}\ \int_a^b (k-h)\; , \end{split}$$ because, using (0), it is easy to verify that: $$\left( \frac{h^\prime}{\sqrt{1+(h^\prime)^2}} \right)^\prime = -\frac{1}{R}; .$$ Therefore, plugging (2) into (1) and keeping in mind that $|D|=\int_a^b (k-h)$, we get: $$\tag{3} \mathcal{L}(k)-\mathcal{L}(h)\geq \frac{1}{R}\ |D|$$ which is the claim.

Moreover equality in (3) holds iff it holds in the convexity inequality; since the function $t\mapsto \sqrt{1+t^2}$ is strictly convex, equality holds iff $k^\prime =h^\prime$ in $(a,b)$, hence $k-h$ is constant. On the other hand $k(a)=h(a)$, therefore the the constant equals zero, i.e. $k=h$ in the whole $[a,b]$.

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I would like to present another proof, which is more like a remark, than an answer, and may provide insight as to where this inequality comes from. It is a well known result that the isoperimetric problems. Consider $\Omega \subset \Bbb{R}^N$ an open set. Then

$$ \min_{E \subset \Omega, |E|=c} Per(E)$$

and

$$ \max_{E \subset \Omega, Per(E)=c} |E|$$

have a solution in $\Omega$, if the parameter $c$ is chosen such that the set of admissible forms is not empty. Moreover, it can be shown that the mean curvature of the optimal shape is constant, on parts of $E_{optimum}$ which lie inside $\Omega$. We can make the class of admissible even smaller, by requiring the optimum to have a fixed part of the boundary $\Gamma \subset \overline\Omega$. (the proof relies on a compactness argument and can be obtained using similar reasoning as here).

So if you take $\Omega=h_+=\{ (x,y): x \in [a,b],y \geq h(x)\}$, and $c=L(k)$, we can maximize $|D|$, and such a maximizer $E_0$, such that the graph of $h$ is included in the boundary of $E_0$ exists in $\Omega$. The fact that the minimizer has mean constant curvature inside $\Omega$ tells us that the boundary of the optimum $E_0$ is an arc of circle inside $\Omega$, which will be a translate of $h$. To satisfy the constraint $L(k)=c\geq L(h)$, the segments which are boundaries of $E_0$ on the lines $x=a$ and $x=b$ must be of length $d=(c-L(h))/2$. In this case, the area of $D$ can be calculated easily, using an integral and it is $2R \times d=R(c-L(h))=R(L(k)-L(h))$.

Therefore, the maximum area of $D$ when the length of $k$ is prescribed is equal to $R(L(k)-L(h))$. This proves that for arbitrary $k$ with prescribed length we have $$ \frac{1}{R}|D|\leq L(k)-L(h)$$ and the equality cannot have place if $k \neq h$ because the optimal form has two vertical segments and cannot be written as a graph of a function with $k(a)=k(b)=h(a)=h(b)$.

Since this result is valid for any choice of $L(k)=c \geq L(h)$, it follows that the given inequality holds, and the equality cannot hold unless $h=k$.

I know that this it is not as rigorous as the first answer, but it may explain in another way why the inequality is true.

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+1 for your good Geometric Measure theoretic answer. –  Pacciu Mar 12 '12 at 2:51
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