# Let $Y$ be a finite-dimensional normed space, $X$ a normed space, and $T: X \to Y$ a surjective linear operator. Show that $T$ is an open mapping.

Let $$Y$$ be a finite-dimensional normed space, $$X$$ a normed space, and $$T: X \to Y$$ a surjective linear operator. Show that $$T$$ is an open mapping.

I think if I can show that $$T(B_X)$$ contains an open ball then I am done where $$B_X$$ is the unit ball in $$X$$. But I am unable to show that. Need some help...

1. If $T$ is continuous, then $\ker(T)$ is closed and the quotient map $$\pi : X \to X/\ker(T)$$ is an open map. Furthermore, $T$ induces an injective map $$S : X/\ker(T) \to Y$$ Since $Y$ is finite dimensional, so is $X/\ker(T)$, and so $S$ (whose range is $Y$) is now a homeomorphism. In particular, $S$ is an open map, so $$T = S\circ \pi$$ is also open.

2. If $\dim(Y) = 1$, then it follows from an earlier question that if $U$ is a non-empty open set, then $T(U) = \mathbb{C}$, so it is, in particular, an open map.

3. Not sure about the general case (if $T$ is discontinuous and $\dim(Y) > 1$), but perhaps someone else can complete that case (I don't think induction works, but perhaps it could)

• If T is continuous then it is the open mapping theorem..Only Interesting case is to prove the result when T is discontinuous. – Saikat Oct 2 '15 at 9:28
• $X$ need not be complete, so open mapping theorem does not necessarily apply. And even if $X$ is complete, the proof above is far easier because the range is finite dimensional. – Prahlad Vaidyanathan Oct 2 '15 at 9:32

Lemma : Let $T$ is an onto linear map from $X$ to $Y$, where $X,Y$ are n.l.s and $Y$ is finite dimensional. Let $U \subseteq X$ is open such that $T(U)$ has only non zero elements.Then $T(U)$ is open.

Proof : Let dimension of $Y$ be a natural no. k. Let $u \in U$ ,then $Tu \ne 0$. Extend $Tu$ to a basis of $Y$ say $\{Tu =: u_1,u_2, .... ,u_k\}$. Clearly $u \ne 0$. Choose $v_i \in X$, such that $Tv_i = u_i \forall i = 1, ..., k$ and $v_1 = u$. Thus $\{v_1,v_2, .... ,v_k\}$ is a linearly independent set in $X$. Let $X_u := span \{v_1,...., v_k\} \le X$. Let $T_u$ denote the restriction of $T$ onto $X_u$. Now $T_u$ : $X_u$ $\rightarrow Y$ is a linear isomorphism (also homeomorphism of topological spaces).Now $U$ $\cap$ $X_u$ is open in $X_u$, therefore $T(U \cap X_u)$ is also open in $Y$. Now $u \in U \cap X_u$ so $Tu \in T(U \cap X_u) \subseteq T(U) \subseteq Y$. Thus $T(U)$ is open in $Y$.

Enough to show $T(B(0,r))$ is open in $Y$ for all $r \gt 0$ (since translations are homeomorphisms). We have assumed the dimension of $Y$ to be a natural no., otherwise the result is vacuous.If $T(B(0,r))$ is $Y$ then we are done, else choose $y \in Y\backslash T(B(0,r))$. Let $t_y$ denote the translation in $Y$ by $y$. Let $x \in X$ such that $Tx = y$. Let $t_x$ denote the translation by $x$ in $X$. (note $y,x$ are non zero vectors) Clearly $T(t_x(B(0,r))=B(x,r)) = t_y(T(B(0,r))) = T(B(0,r)) + y$. We show that the last set do not contain zero : Let $z \in B(0,r)$, then $Tz + y = 0$ implies $T(-z) = y$, therefore $y \in T(B(0,r))$, contradiction (since$-z \in B(0,r)$). Thus the last set is open in $Y$ by the lemma. Now since translations are homeomorphisms, we have $T(B(0,r))$ is open. This completes the proof.
WLOG, we can take $$Y=(\mathbb{C}^{n},\|\cdot\|_{\infty})$$(why?). Let $$\{e_1,e_2,\dots,e_n\}$$ be the standard basis of $$\mathbb{C}^{n}$$. Since $$T$$ is surjective, we can find $$a_1,a_2,\ldots,a_n$$ in $$X$$ such that $$T(a_i)=e_i$$ for all $$1\leq i\leq n$$. Define $$\alpha:=\sum_{i=1}^n\|a_i\|$$, a positive real number. Now, let $$V$$ be an open subset of $$X$$ and consider an element $$x\in V$$. As $$V$$ is open, we can find an $$r> 0$$ such that $$B(x;r)\subseteq V$$.
Consider $$k=(k_1,k_2,\ldots,k_n)\in \mathbb{C}^{n}$$ with $$\|k\|_{\infty}\leq \frac{r}{\alpha}$$. Then $$x-\sum_{i=1}^nk_ia_i\in V$$, so that $$T(x-\sum_{i=1}^nk_ia_i)=T(x)-\sum_{i=1}^nk_ie_i=T(x)-k\in T(V)$$. Thus, the set $$\left\{k\in \mathbb{C}^{n}: \|T(x)-k\|_{\infty}\leq \frac{r}{\alpha}\right\}\subseteq T(V),$$showing that $$T(V)$$ is open in $$Y$$.