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j___d

#1 h Mar 6, 2017, 5:06 PM • 4 Y

Y by Mathuzb, Adventure10, Mango247, mathematicsy

Let $f\in C^1[a,b]$ , $f(a)=0$ and suppose that $\lambda\in\mathbb R$ , $\lambda >0$ , is such that
$|f'(x)|\leq \lambda |f(x)|$ for all $x\in [a,b]$ . Is it true that $f(x)=0$ for all $x\in [a,b]$ ?

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lambda

#2 h Mar 7, 2017, 2:21 AM • 1 Y

Y by Adventure10

j___d wrote:

Let $f\in C^1[a,b]$ , $f(a)=0$ and suppose that $\lambda\in\mathbb R$ , $\lambda >0$ , is such that
$|f'(x)|\leq \lambda |f(x)|$ for all $x\in [a,b]$ . Is it true that $f(x)=0$ for all $x\in [a,b]$ ?

Let $A=\{|f(x)|=0\}$ . $A$ is nonempty and open. $x\in A$ , for $y\sim x\pm\delta$ , we have $|f(y)|\leq 2\lambda\delta M, M=\max|f(x)|$ , repeat it, we have $|f(y)|\leq (2\delta \lambda)^nM$ , if $\delta$ is selected properly then $f(y)=0$ necessarily. $A$ is open!, $A=[a,b]$

This post has been edited 3 times. Last edited by lambda, Mar 7, 2017, 2:22 AM

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#3 h Jul 24, 2017, 9:44 AM • 3 Y

Y by Adventure10, Mango247, bakkune

Set $M= \sup_{x \in [a,b]} \lvert f(x) \rvert$ . Applying the fundamental theorem of calculus, we conclude that for every $x \in [a,b]$ $\lvert f(x) \rvert \leq \int_{a}^{x} \lvert f'(t) \rvert dt \leq \lambda \int_{a}^{x} \lvert f(t) \rvert dt \leq \lambda (x-a) M$ Using this, we can derive in a similar way that
$\lvert f(x) \rvert \leq \int_{a}^{x} \lvert f'(t) \rvert dt \leq \lambda \int_{a}^{x} \lvert f(t) \rvert dt \leq \int_{a}^{x} \lambda^2 (t-a) M dt = M \frac{\lambda^2 (x-a)^2}{2!}$ Iterating this procedure by means of induction, we eventually deduce $\lvert f(x) \rvert \leq M \, \frac{\lambda^{n}(b-a)^{n}}{n!} \rightarrow 0$ as $n\rightarrow \infty$ .

This post has been edited 3 times. Last edited by Oscillator, Aug 8, 2022, 8:51 PM
Reason: Clarification and editing typos

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TomMarvoloRiddle

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TomMarvoloRiddle

#4 h Jul 24, 2017, 1:24 PM • 3 Y

Y by Vimath, Adventure10, Mango247

Consider the function $g (x)=e^{-\lambda x}|f(x)|$
And see that $g'(x)\leq 0$
So $g$ is decreasing.
So, $x\geq a \implies g (x)\leq g(a)=0$
Hence, $|f (x)|\leq 0\implies f (x)=0$

This post has been edited 3 times. Last edited by TomMarvoloRiddle, Jul 24, 2017, 4:24 PM

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#5 h Jul 24, 2017, 4:39 PM • 2 Y

Y by Adventure10, Mango247

TomMarvoloRiddle wrote:

Consider the function $g (x)=e^{-\lambda x}|f(x)|$
And see that $g'(x)\leq 0$
So $g$ is decreasing.
So, $x\geq a \implies g (x)\leq g(a)=0$
Hence, $|f (x)|\leq 0\implies f (x)=0$

Why $|f(x)|$ should be differentiable?

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TomMarvoloRiddle

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TomMarvoloRiddle

#6 h Jul 24, 2017, 5:00 PM • 2 Y

Y by Adventure10, Mango247

Because $f\in C^1 [a,b]$

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2459 posts

#7 h Jul 24, 2017, 5:50 PM • 2 Y

Y by Adventure10, Mango247

So what...? Consider $f(x)=x$ . Apparently $f\in C^1[-1,1]$ but $|f(x)|$ is not differentiable at $x=0$ !?

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#8 h Jul 25, 2017, 7:00 AM • 2 Y

Y by Adventure10, Mango247

Since $|f'(x)|$ is given,so there is no doubt that $|f (x)|$ is differentiable.

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oty

#9 h Jul 25, 2017, 1:10 PM • 1 Y

Y by Adventure10

@above no , $(|f(x)|)'$ is not $|f'(x)|$

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#10 h Jul 25, 2017, 4:43 PM • 2 Y

Y by Adventure10, Mango247

Perhaps a quick application of Gronwall's Inequality might do the trick. We have
$f'(t) \leq \lambda f(t) \ \forall t \in [a,b]$
so by Gronwall's Inequality we know
$f(t) \leq f(a) e^{\int_a ^t \lambda ds} \ \forall t \in [a,b]$
however that just gives that $f(t) \leq 0 \ \forall t \in [a,b]$ .
Since there's an absolute value on both sides, if you consider the other case, the inequality flips and you get that
$f(t) \geq 0 \ \forall t \in [a,b]$
Thus $f(t) = 0$

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Raunit

#11 h Sep 27, 2021, 6:42 PM

Y by

Oscillator wrote:

Set $M= \sup_{x \in [a,b]} \lvert f(x) \rvert$ . Applying the fundamental theorem of calculus, we conclude that for every $x \in [a,b]$ $\lvert f(x) \rvert \leq \lambda \int_{a}^{x} \lvert f(x) \rvert dx \leq \lambda (x-a) M$ Iterating this process by means of induction, we arrive at $\lvert f(x) \rvert \leq M \, \frac{\lambda^{n}(b-a)^{n}}{n!} \rightarrow 0$ as $n\rightarrow \infty$ .

I couldnot understand the iteration part, how to do to it, could you do it for me

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#12 h Aug 8, 2022, 8:53 PM

Y by

Raunit wrote:

Oscillator wrote:

Set $M= \sup_{x \in [a,b]} \lvert f(x) \rvert$ . Applying the fundamental theorem of calculus, we conclude that for every $x \in [a,b]$ $\lvert f(x) \rvert \leq \lambda \int_{a}^{x} \lvert f(x) \rvert dx \leq \lambda (x-a) M$ Iterating this process by means of induction, we arrive at $\lvert f(x) \rvert \leq M \, \frac{\lambda^{n}(b-a)^{n}}{n!} \rightarrow 0$ as $n\rightarrow \infty$ .

I couldnot understand the iteration part, how to do to it, could you do it for me

Sorry for late reply, I edited my previous post slightly, so hope it clarifies!

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#14 h Mar 20, 2025, 5:32 AM

Y by

Oscillator wrote:

Raunit wrote:

Oscillator wrote:

Set $M= \sup_{x \in [a,b]} \lvert f(x) \rvert$ . Applying the fundamental theorem of calculus, we conclude that for every $x \in [a,b]$ $\lvert f(x) \rvert \leq \lambda \int_{a}^{x} \lvert f(x) \rvert dx \leq \lambda (x-a) M$ Iterating this process by means of induction, we arrive at $\lvert f(x) \rvert \leq M \, \frac{\lambda^{n}(b-a)^{n}}{n!} \rightarrow 0$ as $n\rightarrow \infty$ .

I couldnot understand the iteration part, how to do to it, could you do it for me

Sorry for late reply, I edited my previous post slightly, so hope it clarifies!

ahh ofc, it took you 5 years to reply, so early ig it would have better if you didnt :rotfl:

:rotfl:

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