# Map that satisfies $f(\lambda x) = \lambda f(x)$ but not $f(x+y) = f(x)+f(y)$

Could you give me example of maps $f:\mathbb R \to \mathbb R$ that satisfy $$f(\lambda x) = \lambda f(x) \quad \forall x,\lambda \in \mathbb R$$ but not $f(x+y) = f(x)+f(y)$? Thanks in advance.

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– Zev Chonoles Mar 8 '13 at 22:52
$f(\lambda)=\lambda f(1)$ so $f(x+y)=x f(1) + y f(1) =f(x) + f(y)$ – user58512 Mar 8 '13 at 22:53
You'll almost certainly have more success in higher dimensions... – Micah Mar 8 '13 at 23:01

Let $f(1)=a$. Then $f(\lambda\cdot 1)=\lambda f(1)=\lambda a$. Or, to put it in more familiar letters, $f(x)=ax$. So $f(x+y)=f(x)+f(y)$.

Remarks: Things get more interesting if, for example, we restrict $\lambda$ to rational values.

Things also get more interesting if we ask about the converse problem. It turns out that there are exotic solutions to the Cauchy functional equation $f(x+y)=f(x)+f(y)$.

Things also get more interesting if replace $\mathbb{R}$ by $\mathbb{R}^2$. Then it is perfectly possible to have $f(\lambda x)=\lambda f(x)$ for all vectors $x$ and scalars $\lambda$, without having $f(x+y)=f(x)+f(y)$. For example, map all points on the $x$-axis to $(0,0)$, and map off-axis points to themselves.

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Ok, what a trivial statement. Then, why linear maps are defined requiring additivity, which is consequence of homogeneity of degree 1? Why not requiring only that $f(\lambda x ) = \lambda f(x)$? – V. Galerkin Mar 8 '13 at 23:02
From $\mathbb{R}$ to $\mathbb{R}$, doesn't do anything. But there are plenty of mappings from $\mathbb{R^2}$ that satisfy, for vectors $x$ and scalars $\lambda$, the equation $f(\lambda x)=\lambda f(x)$ but are not linear. – André Nicolas Mar 8 '13 at 23:06
@V.Galerkin Because for linear maps, the argument is a vector, not a scalar. So you can't show $T(x + y) = T(x) + T(y)$ using $T(cx) = cT(x)$ (where c is a scalar) since you can not take out the vector $x + y$ to get $(x + y)T(1)$. This worked for you question as the argument was a scalar. – Pratyush Sarkar Mar 8 '13 at 23:29

No. Choose $z$ with $f(z)\ne 0$. Then $f(x+y)=\frac{x+y} zf(z)=\frac xzf( z)+\frac yz f(z)=f(x)+f(y)$

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