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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 $x \in [a, b]$.

  • 2
    This is [Cauchy's mean value theorem](http://en.wikipedia.org/wiki/Cauchy%27s_mean_value_theorem#Cauchy.27s_mean_value_theorem).2011-12-27

2 Answers 2

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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​​].

2

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 $f(x)=0$.

  • 0
    i think yes, because we have not yet addressed integral....2011-12-29