Operators between normed linear spaces Prove or give a counterexample about an operator $T$ from a normed linear space $X$ to another normed linear space $Y$ ($T\colon X\to Y$) which is an additive operator (holds addition homomorphism), that is to say, for any $x\in X$, $T(x+y)=T(x)+T(y)$. And we give an additional condition, “bounded” ($T$ maps a bounded set into a bounded set).
I have known that if $T$ is a linear operator (not only satisfies addition homomorphism but also satisfies scalar multiple homomorphism for any $\alpha\in K\text{ ($R$ or $C$)}$，$x\in X$, $T(\alpha x)=\alpha T(x)$, $T$ is bounded iff $T$ is continuous iff $T$ maps any bounded set into a bounded set.
But now $T$ is just an addition homomorphism and $T$ maps any bounded set into a bounded set, can we get that $T$ is a linear operator. I found it is so difficult that I can’t prove it or give a counterexample. Please give a proof or a counterexample.
 A: I am guessing that you are asking if $T$ is a bounded additive homomorphism then it is linear.
If $T$ is bounded, then it is continuous. Let $\overline{B}$ be the closed unit ball, then for some $K< \infty$ we have $\|T(x)\| \le K$ for all $x \in \overline{B}$. Since $T(x)-T(y) = T(x-y)$, we need only examine continuity at zero. We have $T(nx) = n T(x)$ for $n \in \mathbb{N}$, so if $\|x\| \le \frac{1}{n}$ (equivalently $\|nx\| \le 1$), we have $\|T(x)\| \le \frac{1}{n} K$. Continuity follows.
If the field is real, then it is straightforward to show that $T({n \over m} x) = {n \over m} T(x)$, and continuity shows this is true for all scalars.
Elaboration: $0 = T(x + (-x)) = T(x) + T(-x)$, so $T(-x) = -T(x)$. We have $T(nx) = n T(x)$ for $n \in \{0,1,...\}$, and $T(m (\frac{1}{m} x)) = m T(\frac{1}{m} x)$, which gives $T(\frac{1}{m} x) = \frac{1}{m} T(x)$. Combining shows $T(qx) = qT(x)$ for all $q \in \mathbb{Q}$.
Let $T : \mathbb{C} \to \mathbb{C}$ be given by $T(x) = \overline{x}$. Then $T$ is a bounded additive homomorphism, but is not linear.
If, in addition, $T(ix) = iT(x)$, then it is true with a complex field.
