Is this function constant and is that function differentiable ?

[God]

if |f(x)-f(y)|<=sin²(x-y) for every x and y reals then f is constant. True or false ?

I know that sin²(x-y) is between 0 and 1 , if it's true I'd have |f(x)-f(y)|<=0 but if sin² is equal to 1, I have |f(x)-f(y)|<=1 and I can't conclude for every x and y reals...

the other function is such as g(x)=0 if x is irrationnal and g(x)=x² is x is rationnal
Is this function differentiable in 0 and if so, does it own a relative minimum at that point ?

I think such a function can't be continuous since near 0 there are infinitely small rationnals ( 1/9999999999999) and irrationnals too (sqrt(1/9999999999999999999999999)) so it cannot be differentiable right ?

 

[daon2]

Let y=x+h, for x,y different. Then you have:

|f(x+h)-f(x)| <= sin^2(h)

Now divide by |h|.

See what to do now?

 

[God]

Isnt it sin²(-h) ? how can I prove lim h->0 sin²(-h)/|h|=0 ?
The derivative, which tends to 0 as h tends to 0, so it is true, f is constant ?
it is still weird because let's x-y=pi/2
f(y+pi/2)-f(y)<=1

doesnt seem to be constant...

 

[daon2]

Isnt it sin²(-h) ? how can I prove lim h->0 sin²(-h)/|h|=0 ?
The derivative, which tends to 0 as h tends to 0, so it is true, f is constant ?
it is still weird because let's x-y=pi/2
f(y+pi/2)-f(y)<=1

doesnt seem to be constant...

What do you mean "doesn't seem to be constant"? If f is a constant function, then f(y+pi/2)-f(y)=0.

Second, sin^2(-h)=(sin(-h))^2 =* (-sin(h))^2 = sin^2(h). Where the starred equal sign holds true because sin is an odd function.

Third, you probably already know the limit of sin(x)/x as x->0. If not, then I am not sure how this problem would otherwise be solved.

edit:

I did a google search just now and found a similar question asked with sin^2(x-y) replaced with |x-y|^2. This leads to another solution but requires you to know the identity: sin(x) <= |x|.

Following the original approach, we have \(\displaystyle 0\le \left |\dfrac{f(x+h)-f(x)}{h}\right |\le \dfrac{\sin^2(h)}{|h|} \le \dfrac{|h|^2}{|h|} = |h|\)

 

[HallsofIvy]

if |f(x)-f(y)|<=sin²(x-y) for every x and y reals then f is constant. True or false ?

I know that sin²(x-y) is between 0 and 1 , if it's true I'd have |f(x)-f(y)|<=0 but if sin² is equal to 1, I have |f(x)-f(y)|<=1 and I can't conclude for every x and y reals...

the other function is such as g(x)=0 if x is irrationnal and g(x)=x² is x is rationnal
Is this function differentiable in 0 and if so, does it own a relative minimum at that point ?

I think such a function can't be continuous since near 0 there are infinitely small rationnals ( 1/9999999999999) and irrationnals too (sqrt(1/9999999999999999999999999)) so it cannot be differentiable right ?

So what? If x is a very small irrational, then f(x)= 0 and if x is a very small rational, then f(x)= x^2 which is very close to 0. Why would there be any problem?

 

[God]

Thanks a lot for your help, both questions are solved now.