Question on Mean Value Theorem

[Titan97]

Homework Statement

Let ###f## be double differentiable function such that ##|f''(x)|\le 1## for all ##x\in [0,1]##. If f(0)=f(1), then,
A)##|f(x)|>1##
B)##|f(x)|<1##
C)##|f'(x)|>1##
D)##|f'(x)|<1##

Homework Equations

MVT: $$f'(c)=\frac{f(b)-f(a)}{b-a}$$

The Attempt at a Solution

I first tried using integration.
$$-1\le f''(x) \le 1$$
integrating from 0 to x,
$$-x\le f'(x)-f'(0) \le x$$
Again integrating from 0 to x,
$$-\frac{x^2}{2}\le f(x)-f(0)-f'(0)x \le \frac{x^2}{2}$$
But even though I got an inequality for f(x), I could not remove the constants.
Then I applied Rolle theorem for f(x). Since f(0)=f(1), there exists a point (at least one point) ##c## such that f'(c)=0.
There exists a point ##X\in [c,x]## such that
$$f''(X)=\frac{f(x)-f(c)}{x-c}$$
Here, ##x\in [c,1]##, and since 1>x>c, x-c<1. Also, ##|f''(x)|\le 1##.
$$f''(X)=\frac{f(x)-f(c)}{x-c}$$
So, ##\frac{f(x)-f(c)}{x-c}\le 1## and ##f(x)-f(c)\le {x-c}##. Hence I get the answer D. Is this correct?

 

[pasmith]

Homework Statement

Let ##f## be double differentiable function such that ##|f''(x)|\le 1## for all ##x\in [0,1]##. If f(0)=f(1), then,
A)##|f(x)|>1##
B)##|f(x)|<1##
C)##|f'(x)|>1##
D)##|f'(x)|<1##

Homework Equations

MVT: $$f'(c)=\frac{f(b)-f(a)}{b-a}$$

The Attempt at a Solution

I first tried using integration.

Metahints for real analysis:

If you're given a continuous function on a closed bounded interval and told its values at the end points, consider applying the intermediate value theorem.
If you're given a differentiable function on a closed bounded interval and told its values at the end points, consider applying the mean value theorem.

Then I applied Rolle theorem for f(x). Since f(0)=f(1), there exists a point (at least one point) ##c## such that f'(c)=0.
There exists a point ##X\in [c,x]## such that

$$f''(X)=\frac{f(x)-f(c)}{x-c}$$

There are two problems here. Firstly I take it you are applying the MVT to [itex]f'[/itex] rather than [itex]f[/itex], so [itex]X[/itex] satisfies [tex]
f''(X) = \frac{f'(x) - f'(c)}{x - c} = \frac{f'(x)}{x - c}.[/tex]

Secondly, writing [itex]X \in [c,x][/itex] implicitly requires that [itex]c \leq x[/itex]. But you have to prove a result for every [itex]x \in [0,1][/itex], so you have also to deal with the case [itex]x > c[/itex]. However that requires no further work, since as long as [itex]x \neq c[/itex] there is an [itex]X[/itex] lying between [itex]x[/itex] and [itex]c[/itex] which satisfies the above equation.

Here, ##x\in [c,1]##, and since 1>x>c, x-c<1. Also, ##|f''(x)|\le 1##.

Thus if [itex]x = c[/itex] then [itex]|f'(x)| = 0[/itex], and if [itex]x \neq c[/itex] then [itex]|f'(x)| < \dots[/itex]

 

[RUber]

The Attempt at a Solution

I first tried using integration.
$$-1\le f''(x) \le 1$$
integrating from 0 to x,
$$-x\le f'(x)-f'(0) \le x$$
Again integrating from 0 to x,
$$-\frac{x^2}{2}\le f(x)-f(0)-f'(0)x \le \frac{x^2}{2}$$
But even though I got an inequality for f(x), I could not remove the constants.

What does that inequality look like when x = 1? That will give you bounds on f'(0).

 

[RUber]

What if you define ##f'(x) = f'(0) + \int_0^x f''(t) dt ##?
Then using the inequality ## \left| \int f dx \right| \leq \int |f| dx##
You can quickly deduce something about f'(x).

Then, since f(0) = f(1), you can say that
##\int_0^c f'(x) dx = -\int_c^0 f'(x) dx##
You can find a maximum for this based on the same inequality.

I doubt you can say anything about |f(x)| since you aren't given anything that might constrain the initial values. f(0) = f(1) = 100 might be an option based on what you provided.

 

[Titan97]

@pasmith , that was a typo. I lost internet connection while trying to edit. @RUber, i found the minimum and maximum vaue f'(0) can take. $$-1/2 \le f'(0) \le 1/2$$

 

[Titan97]

But, from the equation, ##-x\le f'(x)-f(0)\le x## and by substituting the max value of f'(0), ##-x+0.5\le f'(x)\le x+0.5## and at x=1, f'(x)>1. Is that correct?

 

[Qwertywerty]

Titan97 , do you want a proper solution , or would getting the answer alone simply be enough ?

I haven't actually tried the question , but you could consider a function as f(x) = x(x - 1)/4 .

Solving this can easily tell you about f'(x) ; however I don't believe you can comment on f(x) on the basis of what's given in the original question .

I know this isn't a proper solution , but still ... Hope this helps .

 

[Titan97]

I want a proper solution. I know many substitutions that can give the answer to such questions.

 

[Qwertywerty]

I want a proper solution. I know many substitutions that can give the answer to such questions.

Okay , first of - You cannot comment on value of f(x) .

For f'(x) - Let f'(x) = k ( f'(0) + x ) , where k is such that -1 <= k <= 1 .

f'(0) = - 0.5 , and so f'(x) = k ( x - 0.5 ) .
( x - 0.5 ) varies from - 0.5 to 0.5 .
⇒ k ( x - 0.5 ) will belong to some interval lesser than equal ( - 0.5 , 0.5 ) - Depending on the value of k

Hope this helps .

 

[Titan97]

what about post 6?

 

[Infrared]

Hint: There is a point [itex] x_0\in (0,1)[/itex] such that [itex]f'(x_0)=0[/itex]. If [itex]x[/itex] is another point in [itex](0,1)[/itex], then you can use the fact that [itex]|x-x_0|<1[/itex] together with the information about [itex]f''[/itex] to estimate [itex]f'(x)[/itex].

 

[Qwertywerty]

what about post 6?

F'(0) has a fixed value -

f'(0) = - 0.5 , and so f'(x) = k ( x - 0.5 ) .
( x - 0.5 ) varies from - 0.5 to 0.5 .

 

[Qwertywerty]

Okay , first of - You cannot comment on value of f(x) .

For f'(x) - Let f'(x) = k ( f'(0) + x ) , where k is such that -1 <= k <= 1 .

f'(0) = - 0.5 , and so f'(x) = k ( x - 0.5 ) .
( x - 0.5 ) varies from - 0.5 to 0.5 .
⇒ k ( x - 0.5 ) will belong to some interval lesser than equal ( - 0.5 , 0.5 ) - Depending on the value of k

Hope this helps .

Made a mistake here . I'll get back to you once I correct it .

 

[RUber]

What if you define ##f'(x) = f'(0) + \int_0^x f''(t) dt ##?
Then using the inequality ## \left| \int f dx \right| \leq \int |f| dx##
You can quickly deduce something about f'(x).

Then, since f(0) = f(1), you can say that
##\int_0^c f'(x) dx = -\int_c^0 f'(x) dx##
You can find a maximum for this based on the same inequality.

I doubt you can say anything about |f(x)| since you aren't given anything that might constrain the initial values. f(0) = f(1) = 100 might be an option based on what you provided.

If you use the facts here, knowing that f'(c)=0, consider that f' goes directly from its initial value to zero as quickly as possible given constraints on f". The smallest c possible is zero. From there, assume max departure from zero using the max for f". How far from zero can f' get on the remainder of the unit interval?

 

[Titan97]

I did not understand that properly @RUber .

 

[Infrared]

I'm not sure why my post #11 is being ignored; it really suffices to solve the problem.

 

[Titan97]

Its the same thing i have given in the original post. It does work.