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 $x \in [a, b]$.

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    $\begingroup$ 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
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    $\begingroup$ This is Cauchy's mean value theorem. $\endgroup$ – Chris Eagle Dec 27 '11 at 19:41

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


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​​].


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 $f(x)=0$.

  • $\begingroup$ 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

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