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Find all functions \(f : \mathbb{R} \rightarrow \mathbb{R}\) such that \(f (xy) = x f(x) + y f(y)\).

Note by Yuxuan Seah 3 years, 2 months ago

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Let \(P(x,y)\) be the statement \(f(xy)=xf(x)+yf(y)\).

\(P(0,0)\implies f(0)=0\)

\(P(x,0)\implies xf(x)=0\)

\(\implies f(x)=\begin{cases}0 && x\neq 0\\ a && a\in\mathbb R, x=0\end{cases}\)

\(P(1,0)\implies a=0\)

Hence \( f(x)=0\) is the only possible solution and by checking it we see it works. \(\square\)

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TopNewestCOOL SOLUTION

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Let \(P(x,y)\) be the statement \(f(xy)=xf(x)+yf(y)\).

\(P(0,0)\implies f(0)=0\)

\(P(x,0)\implies xf(x)=0\)

\(\implies f(x)=\begin{cases}0 && x\neq 0\\ a && a\in\mathbb R, x=0\end{cases}\)

\(P(1,0)\implies a=0\)

Hence \( f(x)=0\) is the only possible solution and by checking it we see it works. \(\square\)

Log in to reply