# Real analysis

## Homework Statement

Mod note: Edited the function definition below to reflect the OP's intent.
Suppose f:R->R is continuous. Let λ be a positive real number, and assume that for every x in R and a>0,f(ax)=aλ f(x). (a) If λ > 1 show that f is differentiable at 0. (b) If 0 < λ < 1 show that f is not differentiable at 0. (c) If λ=1, show that f is differentiable at 0 if and only if it is linear. (Hint: what is f(0)?)

## The Attempt at a Solution

I am considering start the question with f(x)-f(0)/x but how can i find the limit when x approaches 0? f(x)=f(1*x)=1^λf(x) f(0)=f(0*x)=0^λf(x)=0, then is f(x)-f(0)/x = f(x)/x? Then how can i know it's differentiable at 0?

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You must set up a limit, epsilon delta style

You must set up a limit, epsilon delta style
how can i use epsilon delta to prove differentiability?

Mark44
Mentor

## Homework Statement

Suppose f:R->R is continuous. Let λ be a positive real number, and assume that for every x in R and a>0,f(ax)=aλ f(x). (a) If λ > 1 show that f is differentiable at 0. (b) If 0 < λ < 1 show that f is not differentiable at 0. (c) If λ=1, show that f is differentiable at 0 if and only if it is linear. (Hint: what is f(0)?)

## The Attempt at a Solution

I am considering start the question with f(x)-f(0)/x
You are missing parentheses. What you wrote is $f(x) - \frac{f(0)}{x}$. I'm pretty sure you meant $\frac{f(x) - f(0)}{x}$. If you don't use LaTeX like I did, you need to write it as (f(x) - f(0))/x.
silvetriver said:
but how can i find the limit when x approaches 0? f(x)=f(1*x)=1^λf(x) f(0)=f(0*x)=0^λf(x)=0, then is f(x)-f(0)/x = f(x)/x? Then how can i know it's differentiable at 0?
What does ^ mean here? It is often used to indicate exponentiation, as in 2^3 = 8.
Are you using it to indicate "and"?

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You are missing parentheses. What you wrote is $f(x) - \frac{f(0)}{x}$. I'm pretty sure you meant $\frac{f(x) - f(0)}{x}$. If you don't use LaTeX like I did, you need to write it as (f(x) - f(0))/x.
What does ^ mean here? It is often used to indicate exponentiation, as in 2^3 = 8.
Are you using it to indicate "and"?
You are right thats what i mean. yea ^ means exponentiation...

"how can i use epsilon delta to prove differentiability?"

Use the definition of the derivative.

Mark44
Mentor
You are right thats what i mean. yea ^ means exponentiation...
Then I'm really confused now. Where does this come from?
f(x)=f(1*x)=1^λf(x) f(0)=f(0*x)=0^λf(x)=0
I don't see anything in the problem statement about exponents.

In any case, the above is very ambiguous. For example you have 1^λf(x), which I take to mean $1^{\lambda}f(x)$, according to the normal rules for the order of operations. I don't know why you would have $1^{\lambda}$, which equals 1, or $0^{\lambda}$, which equals 0, as you have elsewhere.

Then I'm really confused now. Where does this come from?

I don't see anything in the problem statement about exponents.

In any case, the above is very ambiguous. For example you have 1^λf(x), which I take to mean $1^{\lambda}f(x)$, according to the normal rules for the order of operations. I don't know why you would have $1^{\lambda}$, which equals 1, or $0^{\lambda}$, which equals 0, as you have elsewhere.
oh sorry i mistyped the problem. it should be f(ax)=a^lambda f(x) I think 1^lambda=1 because the problem states that lambda>1 ' of course the a in this definition is not the constant in the definition of f(x)

Mark44
Mentor
oh sorry i mistyped the problem. it should be f(ax)=a^lambda f(x) I think 1^lambda=1 because the problem states that lambda>1
Is it safe to interpret the above as $f(ax) = a^{\lambda}f(x)$?

Is it safe to interpret the above as $f(ax) = a^{\lambda}f(x)$?
yessss i just realize that f(0)=f(a*0)=a^lambda f(0), well then i totally get lost... ' of course the a in this definition is not the constant in the definition of f(x)
yea thats what i think to start with but i can't find f(x) and f(0)

Mark44
Mentor
You must set up a limit, epsilon delta style
The OP has to set up a limit, but I don't believe that a delta-epsilon argument is necessary.

• bubsir
what i think
The nice thing about the definition of derivative AND f(x) is that you only need to find the difference, not the actual function.