# Suppose f : R → R is continuous. Let λ be a positive real number, and assume that for every

Suppose f : R → R is continuous. Let λ be a positive real number, and assume that for every $x ∈ R$ and $a > 0, f(ax) = aλf(x).$

If λ = 1, show that f is differentiable at 0 if and only if it is linear.

I asked a similar question before. But this one is different.

I get $f'(0)=f(1)$ by manipulating the algebra. However, just because $f'(0)=f(1)$ doesn't mean it is linear. So, I need some hints on how to continue the proof.

Besides, since the question asks me to prove "if and only if", which means I'll also have to use "linear" as condition to deduce "$f$ is differentiable at $0$". Any hint on that?

It is another similar question I asked before Differentiability of an homogeneous and continuous function $f$ ($f(\alpha x)=\alpha^\beta f(x)$)

• For our benefit you might want to make the link to the previous question explicit. – Justpassingby Dec 6 '15 at 23:24
• "f'(0) = f(1) doesn't mean it is linear" Um, yes it does.... f(1) is a constant. f is linear if f'(x) = constant. – fleablood Dec 6 '15 at 23:40
• "I'll have to use 'linear' as a contiditon to deduce f is differentiabl at 0. Any hint". That's trivially always true. for any function linear at x, f'(x) = lim f(x+h) - f(x)/h = m(x + h) + b - mx - b/h = mh/h =m exists. – fleablood Dec 6 '15 at 23:42
• With $f'(0) = f(1)"$, I can only show that $f'(x)$ is a constant on 0, not on R. Therefore, I can't conclude that $f(x)$ is a linear function on R – XXWANGL Dec 6 '15 at 23:47
• but f(x) = f(1x) = xf(1). It's linear on x > 0. And andre pointed out for x < 0. f(x) = |x|f(-1) = x(-f(-1)) so it's linear on x < 0. If it's diff at x = 0 then it is continuous -f(-1) = f(1) and f(x) = xf(1) is linear everywhere. – fleablood Dec 6 '15 at 23:55

1) Suppose that $f$ is differentiable at $0$. We show $f$ is linear.

Let $x=1$. Then for every positive $a$ we have $f(a)=af(1)$.

Let $x=-1$. Then for every positive $a$ we have $f(-a)=af(-1)$.

As $a$ approaches $0$ through positive values, $f(a)$ approaches $0$. Also, $f(-a)$ approaches $0$. Thus by continuity we have $f(0)=0$.

The right derivative at $0$ is the limit as $a$ approaches $0$ of $\frac{f(a)-f(0)}{a-0}$, which is $f(1)$.

Similarly, the left derivative is $-f(-1)$. By differentiability, these are equal, so $f(-1)=-f(1)$, and $f(t)=tf(1)$ for all $t$.

2) It is obvious that if the function $f$ is linear, then it is everywhere differentiable.

The 'if' part that your last paragraph refers to is trivial: a linear real function is multiplication by a constant so it is differentiable with derivative the same constant.

For the 'only if' part let $m$ be the derivative at 0. Note that $f(0)=0$ by continuity (start with arbitrary $x$ and divide the argument by increasing powers of 2). We will prove that for arbitrary $x,$ $f(x)=mx.$ In fact we will show that for arbitrary $\epsilon>0,$ $|f(x)-mx|<\epsilon |x|.$ Choose $\delta>0$ so small that for $|t|<\delta,$ $|f(t)-mt|<\epsilon t/2.$ Now set $a=2x/\delta.$ Then

$$|f(x)-mx|=|f(a.\delta/2)-mx|=|af(\delta/2)-am\delta/2|<|a|.\epsilon\delta/2=\epsilon|x|.$$