# Functional

• Nov 7th 2009, 03:54 PM
slevvio
Functional
Hi everyone I was wondering if anybody could help me with this question; I have done the first 2 parts and a seperate 4th part but part 3 is causing me mathematical pain.

Suppose that $\displaystyle f: \mathbb{R} \rightarrow \mathbb{R}$

satisfies $\displaystyle f(t/2) = f(t)/2, t \in \mathbb{R}$.

1) Find f(0).

[easy to show that f(0) = 0 ]

2) Show that $\displaystyle f(2^{-n} t) = 2^{-n}f(t) \forall t \in \mathbb{R} \forall n \in \mathbb{N}$

[can be done easily by induction]

Now suppose that f is differentiable at 0.

3) Show that f(t) = t f'(0) for all $\displaystyle t \in \mathbb{R}$ [Hint: For fixed t, consider the sequence $\displaystyle \{ f(2^{-n}t) / (2^{-n} t) \}$

OK, we know that f is differentiable at 0, so

$\displaystyle \lim_{t \rightarrow 0} \frac{f(t)}{t} = f'(0)$

and for a fixed t, the above sequence simplifies to $\displaystyle \{ \frac{f(t)}{t} \}$ which is a constant sequence, but how does this help me? I guess I can say that

$\displaystyle \lim_{n \rightarrow \infty} \frac{f(t)}{t} = \frac {f(t)}{t}$

Any help would be appreciated :)
• Nov 7th 2009, 04:11 PM
Enrique2
Quote:

Originally Posted by slevvio
Hi everyone I was wondering if anybody could help me with this question; I have done the first 2 parts and a seperate 4th part but part 3 is causing me mathematical pain.

Suppose that $\displaystyle f: \mathbb{R} \rightarrow \mathbb{R}$

satisfies $\displaystyle f(t/2) = f(t)/2, t \in \mathbb{R}$.

1) Find f(0).

[easy to show that f(0) = 0 ]

2) Show that $\displaystyle f(2^{-n} t) = 2^{-n}f(t) \forall t \in \mathbb{R} \forall n \in \mathbb{N}$

[can be done easily by induction]

Now suppose that f is differentiable at 0.

3) Show that f(t) = t f'(0) for all $\displaystyle t \in \mathbb{R}$ [Hint: For fixed t, consider the sequence $\displaystyle \{ f(2^{-n}t) / (2^{-n} t) \}$

OK, we know that f is differentiable at 0, so

$\displaystyle \lim_{t \rightarrow 0} \frac{f(t)}{t} = f'(0)$

and for a fixed t, the above sequence simplifies to $\displaystyle \{ \frac{f(t)}{t} \}$ which is a constant sequence, but how does this help me? I guess I can say that

$\displaystyle \lim_{n \rightarrow \infty} \frac{f(t)}{t} = \frac {f(t)}{t}$

Any help would be appreciated :)

If f is differentiable at 0 then, if we fix t, as $\displaystyle 2^{-n}t$ goes to 0 as $\displaystyle n$ goes to infinity we have

$\displaystyle f'(0)=\lim_{n\to\infty}\frac{f(2^{-n}t)-f(0)}{2^{-n}t}=\lim_{n\to\infty}\frac{2^{-n}f(t)}{2^{-n}t}=\frac{f(t)}{t}$
• Nov 8th 2009, 01:27 PM
slevvio
Is it correct to say this?

$\displaystyle f'(0) = \lim_{t \rightarrow 0} \frac { f(t) - f(0) }{t-0} = \lim_{2^{-n}t \rightarrow 0} \frac { f(2^{-n}t) - f(0) }{2^{-n}t-0} = \lim_{n \rightarrow \infty} \frac { f(2^{-n}t) - f(0) }{2^{-n}t-0}$

I'm getting confused about fixing t. When do we do this, is it at the start? I don't really understand because we are looking what happens when t approaches zero so how can we fix it?
• Nov 8th 2009, 03:54 PM
Enrique2
Quote:

Originally Posted by slevvio
Is it correct to say this?

$\displaystyle f'(0) = \lim_{t \rightarrow 0} \frac { f(t) - f(0) }{t-0} = \lim_{2^{-n}t \rightarrow 0} \frac { f(2^{-n}t) - f(0) }{2^{-n}t-0} = \lim_{n \rightarrow \infty} \frac { f(2^{-n}t) - f(0) }{2^{-n}t-0}$

I'm getting confused about fixing t. When do we do this, is it at the start? I don't really understand because we are looking what happens when t approaches zero so how can we fix it?

No, it is not formally correct. Your problem is that you are mixing t as variable in the definition of derivative and t as a fixed value. Let say, if f is differentiable at 0 then

$\displaystyle f'(0) = \lim_{h \rightarrow 0} \frac { f(h) - f(0) }{h-0}$

This implies that for any zero sequence $\displaystyle (h_n)$ we have

$\displaystyle f'(0) = \lim_{n \rightarrow \infty} \frac { f(h_n) - f(0) }{h_n-0}$

Hence, for a fixed t, you can take $\displaystyle h_n=2^{-n}t$ as a particular zero sequence and apply the hypothesis of differentiability at 0 of $\displaystyle f$ to conclude

$\displaystyle f'(0) = \lim_{n \rightarrow \infty} \frac { f(2^{-n}t) - f(0) }{2^{-n}t-0}$

But as you stated in your post you are "discretisnig" the definition of derivative in a confusing way. For the convergence of the limit of a particular sequence you can't deduce differentiability, you need the existence of the limit for all the possible zero sequences.