Is my logic wrong

Question

On a math.stackechange this question was asked,

Does there exist a function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x+y)=f(x)+f(y)$ for all $x,y \in\mathbb{R}$; $f(1)=1$; and $f(\sqrt 2)=3$?

I solved it as per this,

First of all we start by taking $x,y=0$ $$f(0+0)=f(0)+f(0)$$ $$f(0)=0\cdots (1)$$

Now $$f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$ $$f'(x)=\lim_{h\to 0} \frac{f(x)+f(h)-f(x)}{h}$$ $$f'(x)=\lim_{h\to 0} \frac{f(h)}{h}$$ Now as $f(0)=0$ we can apply the L'Hospital rule, $$f'(x)=f'(0)$$

Now as f'(0) is const, Integrating both the sides, $$\int_{0}^{x} f'(x)dx=\int_{0}^{x} f'(0)dx$$ $$f(x)-f(0)=f'(0)x$$ $$f(x)=f'(0)x$$

Now if, $$f(1)=f'(0)=1$$ It cannot be that, $$f(\sqrt{2})=f'(0)\sqrt{2}=3$$ $$f'(0)=\frac{3}{\sqrt{2}}$$

But one of the answer says such function does exist, where did I go wrong?

Link:$\ref{https://math.stackexchange.com/questions/2428719/does-there-exist-a-function-f-mathbbr→-mathbbr-such-that-fxy-fxfy/2428750#2428750}{ here}$

And if f is non-differentiable, how do we solve these equations?

And can someone name a function if $f$ is non-úmgcos table and satisfies the above conditions?

How can we solve this for non-differentiable functions then?? — neonpokharkar, Sep 14 '17 at 03:56
Indirectly, as per the answer to your other posting of this question. Look at the Wikipedia article https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation . The nonlinear solutions are ugly: unbounded, discontinous, non measurable, and solutions cannot really be exhibited explicitly. — kimchi lover, Sep 14 '17 at 04:00
I did not look closely at your argument, but I believe if you assume $f$ obeys $f(x+y)=f(x)+f(y)$ for all $x$ and $y$, and that $f$ is differentiable at zero, then an argument like yours will deliver $f(x)=ax$ for some $a$. — kimchi lover, Sep 14 '17 at 04:08
But if i take f'(0) RHD and LHD we will get the same answer, so f'(0) must exist? — neonpokharkar, Sep 14 '17 at 04:23
@neonpokharkar See Overview of basic facts about Cauchy functional equation for a discussion of additive functions that are not differentiable (and not linear). — dxiv, Sep 14 '17 at 04:34
Don't forget your differentials ($dx$, $dy$, $d\theta$, etc.) when you integrate. — gen-ℤ ready to perish, Sep 14 '17 at 04:43
From your reference, if f(x) is not differentiable, then it must me not differentiable at all the points? — neonpokharkar, Sep 14 '17 at 04:44
Fallacious. The inference "then you can prove" is obviously not true: the linear solution $f(x)=0 \forall x$ is a counterexample to that step. Fourth line a non-sequitur. — kimchi lover, Sep 14 '17 at 11:25
And why keep it accepted after everybody, its author included, signalled to you that it is incorrect? — Did, Sep 18 '17 at 15:55

kimchi lover · Answer 1 · 2017-09-14T15:19:36.887

The question the OP refers to is this one, which has a concise answer by Robert Israel, which the OP and the "answerer" Isham evidently do not understand. The fact that Cauchy's functional equation has nonlinear solutions has been known for more than a century. (It's a 1905 result of G. Hamel.) All solutions are $\mathbb Q$-linear functions on $\mathbb R$, regarding the latter as a vector space over $\mathbb Q$. Given a Hamel basis for $\mathbb R/\mathbb Q$ one can construct $f$ such that $f(1)=1$ and $f(\sqrt 2)=3$, as Israel shows. But not all $\mathbb Q$-linear functions on $\mathbb R$ are $\mathbb R$- linear, which infuriates the naive. Adding to the infuriating nature of the problem is the fact that there seems to be no easy way to construct and exhibit Hamel bases. But: the original poster's "proof" is fallacious, as is Isham's "answer".

Is my logic wrong

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