linear algebra - existence of continuous non-$1$-homogeneous additive mappings

it's well known that bounded linear maps are continuous $1$-homogeneous additive mappings.

however it doesn't seem trivial to construct a mapping similar to a bounded linear map but without the property of homogeneity of order 1.

I'm well aware that for a sesquilinear form $\xi$ for every fixed $x$ the mapping $y \mapsto \xi_x(x,y)$ is bounded and anti-linear.

but here $\xi_x$ maps from a complex vector space to another.

what if it were a real vector space ? could we still find such a mapping ?

Answer

If $f:V \to W$ is a map between (topological) vector spaces over $\Bbb R$ that satisfies $f(x+y)=f(x)+f(y)$ and $f$ is continuous, then $f$ is $\Bbb R$-linear.

To prove this, note that from additivity we get inductively $f(nx)=nf(x)$ for $n \in \Bbb Z$. Thus if $n \neq 0$ we set $x=\frac{y}{n}$ and divide by $n$, we get $f(\frac{y}{n})=\frac{1}{n}f(y)$. If we combine this with $f(nx)=nf(x)$, we get that for any $q \in \Bbb{Q}$ $f(qx)=qf(x)$.
Now if $r \in \Bbb R$, choose a sequence $q_n$ of rational numbers converging to $r$. Then by continuity we get $f(rx)=f(\lim_{n \to \infty} q_n x)=\lim_{n \to \infty}f(q_nx)=\lim_{n \to \infty}q_nf(x)=rf(x)$, so $f$ is $\Bbb R$-linear.