If a function $f: \mathbb{R}\to\mathbb{R}$ is continuous and $f(x+y) = f(x) + f(y)$ for all $x,y\in\mathbb{R}$, then what is this function $f(x)$?
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2What have you tried? What do you suspect? Which continuous functions can you think of with that property? – Bib-lost May 23 '16 at 14:07
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$f(x)=kx, k \in \mathbb R$ – Roman83 May 23 '16 at 14:10
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Will you please explain how. – Sakshi May 23 '16 at 14:14
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Will it be xf(1) ? – Sakshi May 23 '16 at 14:15
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prove that $f(x) = xf(1)$ step by step, first for $x$ an integer, then for $x$ rational, and finally use continuity of $f$ to conclude this is true for all real numbers $x$. – Josh R May 23 '16 at 14:25
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The first step in solving functional equations is to enter different values for $x$ and $y$ to get simpler expressions that tell you something about the function. For instance, $x = y = 0$ tells us that $f(0+0) = f(0) + f(0)$, from which you can tell that $f(0) = 0$. Also, we may just set $x = y$ to get $f(2x) = 2f(x)$. Now setting $y = 2x$ we get $f(3x) = 3f(x)$. Continuing along these lines (technically by using induction), we get $$f(nx) = nf(x)\tag{1}$$ for all natural numbers $n$ and al real numbers $x$.
Now, using $(\text 1)$ for $x = \frac yn$, we have $f(y) = nf(y/n)$. With rearranging, we get $f(y/n) = f(y)/n$ for any real number $y$ and any natural number $n$. Combining with $(\text 1)$, we get $$ f\left(\frac{m}{n}\cdot x\right) = \frac{m}{n}\cdot f(x) \tag 2 $$ for any real $x$ and rational number $\frac mn$.
Now is the time to use the assumption that $f$ is continuous. Let $k = f(1)$. Then for any rational number $\frac{m}{n}$ equality $(\text 2)$ tells us that $f(m/n) = k \cdot m/n$. Let $x \in \Bbb R$. If $x \in \Bbb Q$, then $f(x) = kx$. If $x$ is irrational, let $x_n$ be a sequence of rational numbers converging to $x$. By continuity of $f$, $f(x_n)$ converges to $f(x)$. But the sequence $f(x_n) = kx_n$, so th sequence converges to $kx$. Therefore we must have $f(x) = kx$ even for irrational $x$.
It is relatively easy to check that any real value $k$ gives a valid solution $f(x) = kx$, and we've shown that any solution is of that form, so we're done.
Arthur
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