Differentiability proof.

[royal condor]

I want to prove that if ( f ) is differentiable at 0, then ( f ) is differentiable on ( \mathbb{R} ), such that for all ( x, y \in \mathbb{R}^2 ), the function satisfies:
[
f(x + y) = f(x) f(y)
]

[silk goblet]

∞ℒ ℳoπitz_ℝeiℕtharDt ℒ∞

[shell cairn]

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[royal condor]

So first, let me tell you what i did.

I stated that for ( f ) to be differentiable at 0, by definition, it must have a limit for ( f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} ). Then, by letting ( y = 0 ) in our function, ( f(x) = f(x) \cdot f(0) ), supposing ( f(x) ) doesn't cancel for all ( x ), then ( f(0) = 1 ). Thus,
[
f'(0) = \lim_{h \to 0} \frac{f(h) - 1}{h}
]

[silk goblet]

∞ℒ ℳoπitz_ℝeiℕtharDt ℒ∞

[royal condor]

I don't know what to do over here, let me think.

should i prove that f is differentiable on R now? and then link the two candidates?

Yea I mean probably

So for ( f ) to be differentiable on ( \mathbb{R} ), it must satisfy
[
f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}.
]
Then, by the functional equation, ( f(x + h) = f(x) f(h) ), so
[
f'(x) = \lim_{h \to 0} \frac{f(x)(f(h) - 1)}{h},
]
since ( f(x) ) is constant when ( h \to 0 ), making
[
f'(x) = f(x) \lim_{h \to 0} \frac{f(h) - 1}{h}.
]

[silk goblet]

∞ℒ ℳoπitz_ℝeiℕtharDt ℒ∞

[royal condor]

We suppose ( f ) is differentiable at 0, then we use the equality we got for
[
f'(0) = \lim_{h \to 0} \frac{f(h) - 1}{h}.
]
Finally, we conclude that
[
f'(x) = f'(0) f(x).
]

[silk goblet]

∞ℒ ℳoπitz_ℝeiℕtharDt ℒ∞

[wide raven]

what does that mean?

you require existence of the limit for differentiability

remove this part

[royal condor]

so it's not exactly true?

[wide raven]

think about it plz..

you are defining differentiability by f'

that is nonsense

[royal condor]

okay so the limit must exist for \lim{h \to 0} \frac{f(x + h) - f(x)}{h}.

[wide raven]

that's better

[royal condor]

Okay

then?

[wide raven]

how do you know this limit exists?

[royal condor]

where? you mean the differentiability at 0?

[wide raven]

$$ \lim _{h\to 0} \frac{f(h) - f(0)}{h} $$

exists

[silk goblet]

aL

[wide raven]

are you saying f(0) = 1?

[royal condor]

yea if f(x) doesn't cancel then yea

[wide raven]

0^2 = 0 is also true

[royal condor]

it is

[wide raven]

so you have no reason to claim f(0) = 1

unless you have some further assumptions

[royal condor]

we have two cases if f(0)= 0 or f(0)=1?

[wide raven]

you have the identity f(x+y) = f(x)f(y)

[royal condor]

i know but should i not assume that the limit $$ \lim _{h\to 0} \frac{f(h) - f(0)}{h} $$
exists

[wide raven]

hence f(0) = f(0) ^2

[silk goblet]

∞ℒ ℳoπitz_ℝeiℕtharDt ℒ∞

[wide raven]

it does exist

but you don't know that f(0) = 1

[royal condor]

how does it exist?

[wide raven]

f is differentiable at zero..

[royal condor]

exactly we have to work with both cases if f(0)=1 or f(0)=0

[royal condor]

correct

[royal condor]

well then the only problem i see is this

[wide raven]

figure out what happens when f(0) = 0

[royal condor]

then f'(0)=limhto0 f(h)/h

right?

[wide raven]

yes

[royal condor]

righttttt

and then what?

[wide raven]

but remember that f(x+y) = f(x)f(y)

[royal condor]

that still doesn't change it

[wide raven]

hint: f(x+0) = f(x)f(0) = 0 for all x

what can you say about f?

[royal condor]

it cancels for this case?

[wide raven]

what does that mean.......

[royal condor]

in terms of differentiability?

[wide raven]

define "function cancels in terms of differentiability"..

https://tenor.com/view/murder-spongebob-rainbow-gif-15431155

[royal condor]

Bro

[wide raven]

what do you say about a function f(x) = 0 for all x?

[royal condor]

constant?

[wide raven]

yes

is that interesting?

[royal condor]

yea lowkey

[wide raven]

no, so f(0) = 1 is the interesting case

if f(0) = 0 then the function is constant and there's nothing to prove

[royal condor]

and then we have to prove when f(0)=1

[wide raven]

you already did

[royal condor]

yea right so i was missing only one case when f(0) = 0 thank you

[wide raven]

try to avoid these nonsense expressions in the future

im being nice about it

your teacher won't be

[royal condor]

I will

be nice about what it was just questions after all.

But thank you

-close

[royal condor]

+close

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