4
$\begingroup$

to solve this problem, it is appropriate to apply Cauchy's theorem?

Let $ h\colon [a,b​​] \to\mathbb R$ a continuous function $ f $ and a differentiable function of $(a,b​)$ such that $ f(a) = 0$. Prove that if there is $ L \neq0$ such that for every $x \in [a, b]$

$$ | L f '(x) + h (x) f (x) | \le | f (x )|,$$

then $f(x)\equiv 0 $ for every $x \in [a, b]$.

$\endgroup$
  • 8
    $\begingroup$ Cauchy had many theorems... $\endgroup$ – David Mitra Dec 27 '11 at 16:31
  • $\begingroup$ I mean this teorem :Let $f, g: [a, b​​] \rightarrow \mathbb {R}$ two real functions of real variable continuous in [a, b], and differentiable on (a, b), with $g ^ {\prime} (x)$ different from $0$ at every point of that range. then $\exists c \in (a, b​​): \frac {f '(c)} {g' (c)} = \frac {f (b) - f (a)} {g (b) - g (a )}.$ Considering in particular the function $g (t) = t $,we obtain the assertion of the theorem of Lagrange. $\endgroup$ – FrConnection Dec 27 '11 at 18:23
  • 2
    $\begingroup$ This is Cauchy's mean value theorem. $\endgroup$ – Chris Eagle Dec 27 '11 at 19:41
3
$\begingroup$

I have tried to resolve the question this way:

the function $h$ is continuous on $[a, b​​]$ then, for the theorem Bolzano (Weierstrass) it's limited, that there exists $A$ such that for every $x\in[a,b​​]$ we have $$|h(x)|\le A.$$

From assumption

$$ | L f '(x) + h (x) f (x) | \le | f (x )|,$$ from which follows

$$ |f '(x)| \le \frac{1+A}{| L|}.$$

Let $[c,d]\subset[a,b]$ of length less than $$\frac{1}{2}\cdot\frac{1+A}{| L|}= \frac{B}{2}$$

and such that

$$f(c)=0$$

We know that in a range with these properties exists (in fact just take for example $c = a$).

Now, if $x_0\in[c, d]$ we can write,

$$|f(x_0)-f(c)|=|f'(x_1)||x_0-c|\le\frac{B}{2}\cdot\frac{|f'(x_1)|}{B}=\frac{|f(x_1)|}{2} \frac{B}{2}$$

Repeating this reasoning we thus find a sequence $(x_n)$ is strictly decreasing and such that

$$f(x_0)\le \frac{|f(x_1)|}{2}\le \frac{|f(x_2)|}{2^2}\le\cdots\le \frac{|f(x_n)|}{2^n}$$

Obviously, this last inequality (here we use the fact that $|f(x_n)|$ is limited) implies $$f(x_0)=0$$ To complete the solution is sufficient to cover $[a, b​​]$ with finite number of subintervals of length less than $\frac{B}{2}$ and use the fact that $f$ is zero on each subinterval. We note that the same conclusion holds if we assume $h$ limited and not necessarily continuous on [a, b​​].

$\endgroup$
2
$\begingroup$

I don't know how to use Cauchy's mean value theorem, but here is a solution. We assume $a=0$ to simplify; we have for all $0\leq x\leq b$ $$|Lf'(x)|\leq |Lf'(x)+h(x)f(x)|+|h(x)f(x)|\leq |(1+h(x))|\cdot|f(x)|,$$ and putting $M:=\sup_{0\leq x\leq b}\left|\frac{1+h(x)}L\right|$, we get $|f'(x)|\leq M|f(x)|$. Now we show by induction that for all $k\geq 1$ and $0\leq x\leq b$: $$\tag{1}|f(x)|\leq \frac{M^k}{(k-1)!}\int_0^x(x-t)^{k-1}|f(t)|dt.$$ For $k=1$, we have, since $f(0)=0$: $$|f(x)|=\left|\int_0^xf'(t)dt\right|\leq M\int_0^x|f(t)|dt,$$ and if it's true for $k$, then \begin{align*} |f(x)|&\leq \frac{M^k}{(k-1)!}\int_0^x(x-t)^{k-1}|f(t)|dt\\ &\leq \frac{M^k}{(k-1)!}\int_0^x(x-t)^{k-1}\int_0^t|f'(s)|dsdt\\ &\leq \frac{M^k}{(k-1)!}\int_0^x(x-t)^{k-1}\int_0^tM|f(s)|dsdt\\ &\leq\frac{M^{k+1}}{(k-1)!}\left(\left[\frac{(x-t)^k}k\int_0^t|f(s)|ds\right]_{t=0}^{t=x}+\int_0^x\frac{(x-t)^k}k|f(t)|dt\right)\\ &=\frac{M^{k+1}}{k!}\int_0^x(x-t)^k|f(t)|dt. \end{align*} Put $M':=\sup_{0\leq x\leq b}|f(x)|$. Then thanks to (1) $$|f(x)|\leq M'\frac{M^kx^k}{k!}\quad \forall k\geq 1$$ so taking the limit $k\to\infty$, we get $f(x)=0$.

$\endgroup$
  • $\begingroup$ good!! but it's possible solve it without use of integrals? $\endgroup$ – FrConnection Dec 29 '11 at 13:50
  • $\begingroup$ It's possible, but I don't know how. Does your exercise requires a proof without integral? $\endgroup$ – Davide Giraudo Dec 29 '11 at 16:47
  • $\begingroup$ i think yes, because we have not yet addressed integral.... $\endgroup$ – FrConnection Dec 29 '11 at 17:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.