# Thread: Proof of symmetric derivative

1. ## Proof of symmetric derivative

Let the symmetric derivative of f at x be,

lim h->0 (f(x+h) + f(x-h) - 2f(x))/(h)

Assume f is continuous on the interval [0,1] and the symmetric derivative exists at all points in (0,1). Prove that there exists a point x in the open interval (0,1) where the ordinary derivative exists.

I've never seen anything like this before. Can anyone help?

Let the symmetric derivative of f at x be,

lim h->0 (f(x+h) + f(x-h) - 2f(x))/(h)

Prove that there exists a point x in the open interval (0,1). Assume the ordinary derivative exists.

I've never seen anything like this before. Can anyone help?
What are you to prove? I assume you meant to say "prove that there exists a point x with some given property". Did you mean to prove that if the ordinary derivative exists at some point then the symmetric derivative exists at that point? Note that (f(x+h)+ f(x-h)- 2f(x))/h= (f(x+h)- f(x))/h+ (f(x-h)- f(x))/h.

3. I edited the question to make it easier to understand.

I edited the question to make it easier to understand.
Follow the hint that HallsofIvy gave you in his post. Notice that the symmetric derivative exists given that the limit as $h \to 0$ of the left hand side of his equation exists. But for that to happen, the limits of the functions on the right must also exist. Note that one is the formula for the ordinary derivative, and the other is a formula equivalent to the formula for the ordinary derivative

5. Alright, here's my first attempt. I sort of understand what I have to do, but I'm not clear on how to properly show it.

Let the symmetric derivative of f at x be,

lim h->0 (f(x+h) + f(x-h) - 2f(x))/(h)

Prove that there exists a point x in the open interval (0,1).

Note that (f(x+h)+ f(x-h)- 2f(x))/h= f(x+h)- f(x))/h + f(x-h)- f(x))/h

Looking at one-sided limits,

Let L = f(x+h)- f(x))/h. Then, the lim h -> 0- f(x+h)- f(x))/h >= 0. As h approaches 0 from the left, the limit must be greater than or equal to zero.

Let R = f(x-h)- f(x))/h. Then, the lim h -> 0+ f(x-h)- f(x))/h <= 0. As h approaches 0 from the right, the limit must be less than or equal to zero. But since we know that f is differentiable at 0, left and right hand limits must coincide. So, 0 <= L = R <= 0.

6. Alright, I attempted the proof again, but I still need to some help. Am I getting closer with this proof? Can anyone help? I kind of need this.

Alright, here's my first attempt. I sort of understand what I have to do, but I'm not clear on how to properly show it.

Let the symmetric derivative of f at x be,

lim h->0 (f(x+h) + f(x-h) - 2f(x))/(h)

Prove that there exists a point x in the open interval (0,1).

Note that (f(x+h)+ f(x-h)- 2f(x))/h= f(x+h)- f(x))/h + f(x-h)- f(x))/h

Looking at one-sided limits,

Let L = f(x+h)- f(x))/h. Then, the lim h -> 0- f(x+h)- f(x))/h >= 0. As h approaches 0 from the left, the limit must be greater than or equal to zero.
I have no idea why you are saying it "must be greater than or equal to zero". For example, if f(x)= 1- x, then f(x+h)= 1- x- h so (f(x+h)-f(x))/h= -h/h which has limit -1. Whether that limit is positive or negative clearly depends upon the function f and you are told nothing about f except that it is differentiable.

You don't need that those limits are positive, or negative, just that they exist and that is true because f is differentiable.

Let R = f(x-h)- f(x))/h. Then, the lim h -> 0+ f(x-h)- f(x))/h <= 0. As h approaches 0 from the right, the limit must be less than or equal to zero. But since we know that f is differentiable at 0, left and right hand limits must coincide. So, 0 <= L = R <= 0.

8. So I just need to show that limits go to something, and that proves there is a point in (0,1) where the ordinary derivative exists?

So I just need to show that limits go to something, and that proves there is a point in (0,1) where the ordinary derivative exists?
yes. but you don't even have to show what they go to. just remember, since $f$ is symmetrically differentiable on (0,1), $\lim_{h \to 0} \frac {f(x + h) + f(x - h) - 2f(x)}h$ exists on (0,1). that tells us something about the first equation HallsofIvy gave

10. We don't know that f is differentiable, though.

We don't know that f is differentiable, though.
yes, i know that. but notice that the symmetric derivative can be written in terms of the ordinary derivative as HallsofIvy showed you. if one side of the equation exists, it forces the other side to exist.

12. Originally Posted by Jhevon
yes, i know that. but notice that the symmetric derivative can be written in terms of the ordinary derivative as HallsofIvy showed you. if one side of the equation exists, it forces the other side to exist.
So, as long as one side exists, similar to what HallsofIvy showed with it being equal to -1, the other side must also exist, regardless of what it is equal to (do I even have to show it equals anything)?

you have that $\lim_{h \to 0} \frac{f(x + h) + f(x - h) - 2f(x)}h = \lim_{h \to 0} \left[ \frac {f(x + h) - f(x)}h + \frac {f(x - h) - f(x)}h \right]$