Does a Point Exist in (0,1) Where the Ordinary Derivative Exists?

[tracedinair]

Homework Statement

Let the symmetric derivative of f at x be

lim h->0 f(x+h) + f(x-h) - 2f(x) / h, provided the limit exists.

Prove there exists a point, x, in (0,1) where the ordinary derivative exists.

Note: f is cont. on [0,1], and the symm. deriv. exists everywhere on (0,1). Prove there is a point, x, in (0,1) where the ordinary derivative exists. Assume f(0)=f(1)=0 and it is unknown if f is differentiable.

Homework Equations

The Attempt at a Solution

I don't know where to start on this. I tried MVT and Rolle's Thm, but unfortunately they don't seem to apply here. Also tried setting h(x) = f(x) - f(0) - x[f(1)-f(0)] and letting g(1)=g(0)=0, but no progress there either. Thx for any help.

 

[mighty2000]

Thank you for your post. This is an interesting problem and I can see why you are having trouble approaching it. Here are some ideas that may help you solve the problem:

1. Consider using the Mean Value Theorem for integrals. This theorem states that if a function f is continuous on [a,b] and differentiable on (a,b), then there exists a point c in (a,b) where the definite integral of f from a to b is equal to f(c)(b-a). This may be helpful in proving the existence of a point where the ordinary derivative exists.

2. You can also try using the Intermediate Value Theorem. This theorem states that if a function f is continuous on [a,b], then for any value k between f(a) and f(b), there exists a point c in (a,b) where f(c)=k. This may help in proving the existence of a point where the ordinary derivative exists.

3. Another approach could be to use the definition of the symmetric derivative. You can try to show that if the symmetric derivative exists everywhere on (0,1), then there must exist a point where the ordinary derivative also exists. This may require some algebraic manipulation and use of the properties of limits.

I hope these ideas help you in solving the problem. Good luck!