Is the function f(t) continuous at t=0?

[pivoxa15]

Is

f(t)= {0, -2<t<0
{t, 0<=t<2

continous everywhere in -2<t<2. I am a bit concerned at t=0

I think yes, after applying the epsilon delta argument but I rememer that a similar function

f(t) = {-t, -2<t<0
{t, 0<=t<2

is not differentiable hence not continuous at t=0.

Have I made a mistake somwhere?

 

[matt grime]

It is perfectly possible for a function to be continuous and not differentiable.

 

[pivoxa15]

So

f(t)= {0, -2<t<0
{t, 0<=t<2

is continuous at t=0?

 

[HallsofIvy]

What is $\lim_{t\rightarrow0}f(t)$?

 

[pivoxa15]

I have to say it is 0 because from the left, it must be 0 since the function is 0. From the right, it is 0 since the function is t. f(0)=0 is defined so it should be continous. But is it differentiable at t=0?

This means
f(t) = {-t, -2<t<0
{t, 0<=t<2
is continuous as well. But it is not differentiable at t=0 is it? How come?

 

[HallsofIvy]

Yes, the limit from both left and right is 0 so the limit is 0. Since the limit there exists and is equal to the value of the function, the function is continuous at 0.

Your f(t) is, of course, |t| between -2 and 2. It is well know that |x|, while continuous, does not have a derivative at x= 0. Essentially, it is not "smooth": there is a corner at x= 0 so the tangent line there is not well defined. More precisely, if h< 0 then f(0+h)= f(h)= -h so
(f(0+h)- f(0))/h= -h/h= -1 while if h> 0, (f(0+h)- f(0))/h= h/h= 1. The two one-sided limits are different so the limit itself, and therefore the derivative at t=0, does not exist.

It's easy to see that the derivative of f is 1 for t> 0 and -1 for t< 0. While the derivative of a function is not necessarily continuous, it does satisfy the "intermediate value property" so the derivative cannot exist at 0.

 

[pivoxa15]

Yes, the limit from both left and right is 0 so the limit is 0. Since the limit there exists and is equal to the value of the function, the function is continuous at 0.

Your f(t) is, of course, |t| between -2 and 2. It is well know that |x|, while continuous, does not have a derivative at x= 0. Essentially, it is not "smooth": there is a corner at x= 0 so the tangent line there is not well defined. More precisely, if h< 0 then f(0+h)= f(h)= -h so
(f(0+h)- f(0))/h= -h/h= -1 while if h> 0, (f(0+h)- f(0))/h= h/h= 1. The two one-sided limits are different so the limit itself, and therefore the derivative at t=0, does not exist.

It's easy to see that the derivative of f is 1 for t> 0 and -1 for t< 0. While the derivative of a function is not necessarily continuous, it does satisfy the "intermediate value property" so the derivative cannot exist at 0.

So |x| is everywhere continuous but not differentiable at x=0 because the derivative function is undefined at x=0.

Back to
f(t)= {0, -2<t<0
{t, 0<=t<2

If I wanted to represent this function as a Fourier series than I am assured that the Fourier series is uniformly convergent in -2<t<2 since f(t) is continuous everywhere in this domain as have been shown. Or could this not happen because if an infinite series of functions is uniformly convergent than the function it represents is continuous everywhere in the domain. But the function being continuous everywhere dosen't always mean uniform convergence of the series representing it. In other words is it 'if' or 'if and only if'?