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Let $f:\mathbb{R}\to \mathbb{R}$ be a function such that $f(x+y)=f(x)+f(y)$ where $x,y\in \mathbb{R}$. If $f(x)$ is differentiable at $x=0$, then

$f(x)$ is differentiable only in a finite interval containing zero

$f’(x)$ is constant for all $x\in \mathbb{R}$

$f(x)$ is constant for all $x\in \mathbb{R}$

$f(x)$ is differentiable except at finitely many points

To pick the right option, the function must be determined first. However, I have not been able to do from the given data, not even a bit. I need a hint to begin the solution process

Multiple correct

Aditya
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  • You've made a (mistaken) assumption when you say, "To pick the right option, the function must be determined first." One possibility is to find some function that satisfies the hypothesis and see whether it rules out any of the four choices. – John Hughes Aug 11 '20 at 12:51
  • This is a classic exercise. The equation is called Cauchy's functional equation. I believe it has been already solved somewhere on this website. – C_M Aug 11 '20 at 12:52
  • Solving Cauchy's functional equation and the solution Clement pointed to are both overkill for this relatively easy exercise. – John Hughes Aug 11 '20 at 12:55
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    Question: It's not clear whether this is a multiple-choice question (i.e., exactly one answer is supposed to be picked) or a question where there might be multiple correct answers (or none), and you're supposed to identify all the correct ones. Can you clarify, please, as it influences how challenging the problem is. – John Hughes Aug 11 '20 at 12:57
  • What's $R$ here? – Bernard Aug 11 '20 at 13:00
  • @Bernard I guess he means $\mathbb{R}$ by $R$ – Ralph Clausen Aug 11 '20 at 13:07
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    @RalphClausen: I guessed so too, but users of the site should learn the basics of MathJax/LaTeX. – Bernard Aug 11 '20 at 13:12
  • That's true @Bernard. – Ralph Clausen Aug 11 '20 at 13:13
  • @JohnHughes it’s multiple correct – Aditya Aug 11 '20 at 14:36

1 Answers1

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Well $f(x + y) = f(x) + f(y) $ for any $x, y \in \mathbb{R}$. Then $f(x) = cx ~$ for any $x \in \mathbb{Q}.$ Now $f$ is differentiable at $x = 0$, i.e., $f(x)$ is cont at $x = 0$.

$\boxed{\blacksquare~ \mathbf{EDIT:}~ \text{The map } f \text{ is in } \mathscr{C}^{0}(\mathbb{R}), \text{ this can be proved form the continuity of } f \text{ at any single }x_0 \in \mathbb{R}, \text{ here }0 = x_0}$

(It's Cauchy's Functional Equation indeed and it's solution mechanism is standard i.e., $$\mathbb{N} \to \mathbb{Z} \to \mathbb{Q} \xrightarrow{\text{if } f ~\in ~\mathscr{C}^0 } \mathbb{R}$$ which is quite trivial)

Then for any given $\epsilon > 0$, $\exists~ \delta > 0$ such that $$ \lvert f(x) - f(0) \rvert < \frac{\epsilon}{2} \quad \text{when }~ \lvert x \rvert < \frac{\delta}{2} $$ Then for any arbitrary $y \in \mathbb{R}$, we have that $$ \lvert f(x) - f(y) \rvert \leqslant \lvert f(x) - f(0) \rvert + \lvert f(y) - f(0)\rvert < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon $$ When $$ \lvert x - y\rvert \leqslant \lvert x \rvert + \lvert y \rvert < \frac{\delta}{2} + \frac{\delta}{2} = \delta $$ Hence $f$ is continuous for any $y \in \mathbb{R}$.

Now it's equivalent to Cauchy's Functional Equation and hence solution is $\color{red}{f(x) = cx,} \color{blue}{\text{ for any } x \in \mathbb{R}} $

Then $f(x) \in \mathscr{C}^1[\mathbb{R}] \implies$ only the $2$nd option is true. $\blacksquare$.

Regrets: This maybe is an overkill of the simple problem. Just note the steps

  • $f$ satisfies the given functional equation and $f$ is cont at some $x_0 \in \mathbb{R} \implies f \in \mathscr{C}^0(\mathbb{R})$.

  • $f(x + y) = f(x) + f(y)$ for any $x, y \in \mathbb{R}$ and $f \in \mathscr{C}^0$ $\implies f(x) = cx ~\forall~ x \in \mathbb{R}$.

This is enough for the problem.

Ralph Clausen
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