# Prove $\int_a^b f(x)dx \leq \frac{e^{2L\beta}-1}{2L\alpha}\int_c^d f(x)dx$

Assume $f(x)>0$ defined in $[a,b]$, and for a certain $L>0$, $f(x)$ satisfies the Lipschitz condition $|f(x_1)-f(x_2)|\leq L|x_1-x_2|$.

Assume that for $a\leq c\leq d\leq b$,$$\int_c^d \frac{1}{f(x)}dx=\alpha,\int_a^b\frac{1}{f(x)}dx=\beta$$Try to prove$$\int_a^b f(x)dx \leq \frac{e^{2L\beta}-1}{2L\alpha}\int_c^d f(x)dx$$

• Is this homework? It's stated as if it were something akin to it. – cardinal Apr 29 '12 at 3:25

## 2 Answers

I got something which is rather close to your result but couldn't get rid of an additional term. I'm hoping someone will find the development useful in order to give a complete answer. The question reminded me somehow of the proof of Gronwall's inequality and my answer is based on that.

Let $h(t)=\int_a^t f(s)ds$ taking the derivative with respect to $t$ yields $h'(t) = f(t)$. Rewriting $f(t)=f^2(t)/f(t)$, we can find an upper bound for $f^2(t)$ indeed, we have: $$f^2(t) = \int_a^t (f^2(s))'ds + f^2(a) = 2\int_a^t f(s)f'(s)ds +f^2(a)$$

but then, the Lipschitz continuity implies that $|f'(s)|\le L$ which yields

$$f^2(t) \le 2L \int_a^t f(s)ds + f^2(a)\quad \Longrightarrow \quad h'(t) \le \left(2Lh(t) +f^2(a)\right)\left({1\over f(t)}\right)$$ we can then introduce the additional positive term $\left(\int_c^df(t)dt\right)/\alpha$ with: $$h'(t) \le \left(2L h(t)+f^2(a) + {\int_c^d f(s)ds\over \alpha}\right)\left({1\over f(t)}\right)$$ Remark: the additional term comes from the $f^2(a)$. We could get rid of it if $\left(\int_c^d f(s)ds\right)/\alpha > f^2(a)$ which I wasn't able to prove.

To simplify notations, let $a=f^2(a) + {\int_c^d f(s)ds\over \alpha},$ $b=2L$ and $g(t)=1/f(t)$ then. Then the previous inequality reads $$h'(t) \le g(t)(a+bh(t))$$ which can be rewritten as follows (here is where it starts to look like Gronwall's inequality):

$${(a+bh(t))'\over a+bh(t)} \le bg(t)$$

the left-hand side is the logarithmic derivative of $a+bh(t)$, integrating both sides from $a$ to $t$ yields

$$a+bh(t) \le a\exp\left(b\int_a^t g(t) \right)$$

plugging the values of $a$,$b$ and using $\int_a^b g(t) = \beta$, we finally get:

$$\int_a^b f(s)ds = h(b) \le \color{green}{\left[{\exp(2L\beta)-1\over 2L\alpha}\right]\int_c^d f(s)ds} + \color{red}{{f^2(a)\over 2L}\left(\exp(2L\beta)-1\right)}$$

• 1.h′(t)=f(t).2.The main problem is that I don't assume that f is derivable.So you may have to improve your method.Thanks. – 89085731 Apr 30 '12 at 0:39
• Yes but I believe that the Lipschitz continuity of $f$ implies that $f$ is differentiable almost everywhere (I believe that this is called Lebesgue's theorem) and that $$\left(f(b)-f(a)\right)=\int_a^b f'(t)dt$$ I agree however that the development can certainly be made more rigorous but the idea was rather to suggest a possible method. – tibL Apr 30 '12 at 7:26
• Since it is a bounty question,I hope that you can show me a rigorous proof. Thanks. Anyway, I appreciate your help. – 89085731 Apr 30 '12 at 10:09

If $f(x)$ is smooth, or $f\in C^1$, assume $\displaystyle h(t)=\int_a^t f(s)\mathrm{d}s$, and $\displaystyle g(t)=\int_a^{t}\frac{1}{f(s)}\mathrm{d}s$,we just can focus on \begin{equation} \frac{h(t)}{\exp(2Lg(t)-1)} \end{equation}

because we know that there is a $\xi\in(c,d)$ s.t. \begin{equation} \frac{h(d)-h(c)}{g(d)-g(c)}=\frac{h'(\xi)}{g'(\xi)}=f^2(\xi) \end{equation}

thus we just prove $$\frac{h(t)}{\exp(2Lg(t)-1)}\le\frac{\min_{[a,b]}f^2(\xi)}{2L}$$

To find the minimum. Let's compute the derivative and we also can see the LHS function at $t=a$ has a limit as $f^2(a)/2L$.

Assume the LHS term is $\gamma(t)$, then \begin{equation} \gamma'(t)=\frac{h'(\exp(2Lg)-1)-2Lg'\exp(2Lg)h}{(\exp(2Lg)-1)^2} \end{equation}

It is easy to compute the limit at $t=a$, we find that $\gamma'(a)=\frac{f(a)f'(a)}{2L}-\frac{1}{2}f(a)\le0$. And we shall see that if we set \begin{eqnarray} p(t)&=&f\cdot(h'(\exp(2Lg)-1)-2Lg'\exp(2Lg)h)\\ &=&f^2(\exp(2Lg)-1)-2L\exp(2Lg)h \end{eqnarray}

We can see that $p(a)=0$, since $h(a)=0$ and $g(a)=0$. However, \begin{eqnarray} p'(t)&=&2ff'(\exp(2Lg)-1)-4L^2\exp(2Lg)h/f\\ &\le&\frac{2L}{f}(f^2(\exp(2Lg)-1)-2L\exp(2Lg)h)\\ &=&\frac{2L}{f}p(t) \end{eqnarray}

Thus we shall know that $p(t)\le 0$, since $\{(\exp(-2Lg(t))p\}'\le0$,and $p(a)=0$.

So we shall get that $\gamma'(t)\le 0$.

Thus $\gamma$ is decreasing in $t$, $\gamma(t)\le\gamma(a)$.

On the other hand. Take $\displaystyle\int_{t}^b f(s)\mathrm{d}s=\phi(t)$,$\displaystyle\int_{t}^b\frac{1}{f(s)}\mathrm{d}s=\psi(t)$.

Then we also can see that \begin{equation} \beta(t)=\frac{\phi(t)}{\exp(2L\psi(t)-1)} \end{equation} which has $\displaystyle\beta(b)=\frac{f^2(b)}{2L}.$

with the same process(PAY ATTENTION TO THE SIGN), $$\beta'(b)=\frac{f(b)f'(b)}{2L}+\frac{f(b)}{2}\ge 0.$$

And also we can obtain that $$q(t)=2L \exp(2L \psi)\phi-f^2(\exp(2L\psi)-1)\ge 0$$

which means $\beta(t)$ is increasing in $t$, which is $\beta(t)\le\beta(b)$.

Since $\beta(a)=\gamma(b)$, thus $\gamma(b)\le \min(f^2(a),f^2(b))/2L$.

AND the choice of $b$ is arbitrary, we know that for any $t$, we have $$\gamma(t)\le f^2(t)/2L$$

Consider the $\min_{[a,b]}f^2(t)$ is reached at $t=\upsilon$, then $\gamma(\upsilon)\le f^2(\upsilon)/2L$, since $\gamma(t)$ is decreasing in $t$. Thus $\gamma(b)\le f^2(\upsilon)/2L$. $\Box$

• When $f$ is Lip. I think one can apply convolution to it to make it smooth. And we can approximate both sides. – Yimin May 3 '12 at 21:57
• 1. I don't understand how you obtain the inequality $\gamma(t) \leq \frac{f(t)^2}{2L}$ "since the choice of $b$ is arbitrary". When $t$ varies, $\gamma(a)$ stays $\gamma(a)$. – Ewan Delanoy May 4 '12 at 5:13
• 2. How do you obtain $\gamma(b) \leq \frac{f(v)^2}{2L}$ from $\gamma(v) \leq \frac{f(v)^2}{2L}$ ? You can't just replace $\gamma(v)$ with $\gamma(b)$. – Ewan Delanoy May 4 '12 at 5:15
• for your last question, since $\gamma$ is decreasing. I proved one thing the integral. If you consider $a$ and $b$ are both variables, then $\gamma(a,b,t)\le\min(\gamma(a,b,a),\gamma(a,b,b))$ – Yimin May 4 '12 at 17:13
• You shall see the definition of $\gamma$, it is $h(t)/\exp(2Lg(t)-1)$, in which, $h$ and $g$ are integrals from $a$ to $t$. thus if for $b\ge a$, we have $h(t)/\exp(2Lg(t))-1)\le f^2(b)/2L$,for any $t\in[a,b]$, we should have the same thing if we consider $b$ as a variable. – Yimin May 4 '12 at 17:18