We know that, if $T$ is a linear operator between normed spaces $X$ and $Y$, then $T$ is continuous $\iff$ $T$ is bounded.
I was wondering if the result holds when $T$ is a sublinear operator. It seems that the proof used for linear operators still works when you use a sublinear operator, yet in every book I read boundedness doesn't seem to be used as synonymous for continuity, and this sounds strange. So I thought maybe I'm not seeing something.
Any help will be appreciated!
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$\begingroup$The conditions on sublinearity imply$$ \left| \ \|T(x_1)\| - \|T(x_2)\|\ \right| \le \|T(x_1-x_2)\|. $$To see this, use $(f,g):=(x_1-x_2,x_2)$ and $(f,g):=(x_2-x_1,x_1)$in those conditions. So if $T$ is bounded then it is continuous at zero. In addition, $x_n\to x$ implies $\|Tx_n\|\to \|Tx\|$. But not much more.
With the following observation it is quite easy to construct a sublinear function: If there is a linear and continuous $L$ such that $\|Tx\|=\|Lx\|$ for all $x$ then $T$ is sublinear and bounded.
Hence the following mapping $T:\mathbb R\to \mathbb R$ is bounded, sublinear, discontinuous:$$ T(x) = \begin{cases} x & \text{ if } x\ne 1\\ -1 & \text{ if } x=+1.\end{cases}$$
Another class of sublinear functions into $\mathbb R$ is given by the Minkowski gauge functional of convex sets.
$\endgroup$ $\begingroup$Let $T:X\to Y$ be a sublinear operator between normed spaces, i.e. $\|T(x+y)\|\leq\|Tx\|+\|Ty\|$ and $\|T(\lambda x)\|=|\lambda\|Tx\|$. Note that any sublinear operator maps $0$ to $0$: we have that $0=\lambda\cdot0$ for all scalars so $\|T0\|=\|T(\lambda 0)\|=|\lambda|\|T0\|$ for all scalars, so $\|T0\|=0$.
Assume that $T$ is continuous. Then $T$ is continuous at $0$, so if $\varepsilon>0$ there exists $\delta>0$ so that $\|x\|<\delta\implies\|Tx\|<\varepsilon$.
Apply this for $\varepsilon=1$ and take a $\delta>0$ satisfying the above implication. Now if $x\in X$ is any non-zero vector then $\frac{\delta}{2\|x\|}\cdot x$ has norm smaller than $\delta$, so $\|T(\frac{\delta}{2\|x\|}x)\|<1$. By the second property in the definition of a sublinear operator we have that$$\frac{\delta}{2\|x\|}\|Tx\|<1$$so $\|Tx\|\leq\frac{\delta}{2}\|x\|$ and therefore $T$ is bounded. Note that we have only used the second property of the definition to prove this direction.
For the converse however, I cannot see how the proof for linear operators can be mimicked for sublinear ones: to show continuity one must estimate $\|Tx-Tx_0\|$ in terms of $\|x-x_0\|$. In the linear setting, $\|Tx-Tx_0\|=\|T(x-x_0)\|$ and boundedness can be used to immediately derive continuity at $0$, so when $\|x-x_0\|$ becomes small we have that $x-x_0\to0$, so $\|T(x-x_0)\|\to0$. However the crucial equality $\|Tx-Tx_0\|=\|T(x-x_0)\|$ does not hold in the sublinear setting and I cannot see an "over the counter" inequality to be used here for our estimation. Unless I am missing something obvious; I have not been involved with sublinear operators so I am not the best person to speak on this.
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