# Problem based on Open Mapping Theorem in Functional Analysis

Let $V$ and $W$ be two Banach spaces and let $T\in L(V,W)$ be bounded such that $R(T)$ is closed and dim $N(T)<\infty$. Let $|.|$ denote another norm on $V$ with $|x|\leq M\|x\|_V$ for all $\,x\in V$, and $\,M$ being a positive constant.

I'd like to prove that there exists a constant $C$ such that $$\|x\|_V\leq C\big(\|Tx\|_W+|x|\big)\quad\text{for all }x\in V.$$

My effort:

Consider $Id:(V,||.||_V)\rightarrow (V,|.|)$ this map is continuous (as given $|x|\leq M\|x\|_V$) and onto so by open mapping theorem $Id$ is open map so there will be a constant $M'$ such that $||x||_V\leq M'|x|$ that means two norms are equivalent. Also given $R(T)$ is closed in banach space $W$, hence $R(T)$ itself banach.

Now define a new norm on $V$ by $||x||_N:=||Tx||_W+|x|$

I check that under this norm $(V,||.||_N)$ is again Banach space. Now finally consider $Id:(V,||.||_V)\rightarrow (V,||.||_N)$ and we obtain

$||Tx||_W+|x|\leq(M+||T||)||x||_V$ for all $x\in V$ So again by open mapping theorem $(V,||.||_V)$ and $(V,||.||_N)$ are equivalent and we will obtain the require result.

Also I observed that there is no requirement of $N(T)<\infty$, Please verify my approach If it is correct then may I put it as answer? Any help/hint in this regards would be highly appreciated.

• Is $T$ bounded? – mechanodroid May 31 '18 at 15:27
• Yes, $T$ is bounded linear operator from $V$ to $W$. – RipCheck May 31 '18 at 15:28

The fact that $N(T)$ is finite dimensional space is needed to show that $\|x\|_N$ is complete:

Consider $X=R(T)\times N(T)$ endowed with a product norm. (Take any norm on $N(T)$ since it is finite dimensional). Let $e_1,...,e_n$ be a basis of $N(T)$, consider $b_i$ the form defined on $N(T)$ by $b_i(e_i)=1, b_i(e_j)=0, i\neq j$. Hahn Banach, you can extend it to $a_i$ on $V$, define $U:V\rightarrow X$ by $U(x)=(T(x),a_1(x)e_1,...,a_n(x)e_n)$. $U$ is bijective and continuous. To see this let $L$ be a $\cap_i Ker(a_i)$, $Im(T_{\mid L}=R(T)$.

Let $(x_n)$ be a $\|\|_N$ Cauchy sequence, $(T(x_n))$ is a Cauchy sequence, it implies that $u_n=U^{-1}(T(x_n),0)$ is a Cauchy sequence of $(V,\|\|_V)$, $T(x_n-u_n)=0$, we deduce that it is a $\|\|_N$ Cauchy sequence in a finite dimensional space $(N(T)$) so it converges, since $u_n$ converges for $\|\|_V$ (since $T(x_n)$ converges and $U^{-1}$ is continuous), it converges for $||$ since $|v|\leq M\|v\|_V$, we deduce that $x_n=(x_n-u_n)+u_n$ converges for $\|\|_N$.

• what happen when $T$ will not be $0$? – RipCheck May 31 '18 at 15:17
• I think that there are conditions that you don't need, it works for every $T$. – Tsemo Aristide May 31 '18 at 15:20
• You are using that $V$ is also complete with respect to the new norm. This is not stated in the question. – Jan Bohr May 31 '18 at 15:42

$T$ gives rise to a bounded linear isomorphism $V/N(T) \rightarrow R(T)$, which has a bounded inverse by the open mapping theorem, i.e. there is some $C_1>0$ with

$$C_1 \Vert Tx \Vert_W \ge \Vert [x] \Vert_{V/N(T)} = \inf_{y\in N(T)}\Vert x- y \Vert_V.$$ Fix some $x$ and consider $f_x\colon N(T)\rightarrow \mathbb{R},y\mapsto\Vert x- y \Vert_V$. Note that $$f_x(y)\ge M^{-1} \vert x-y\vert\ \ge M^{-1}\vert(\vert x \vert - \vert y \vert)\vert,$$ in particular there is a constant $C_2$ such that $\vert y\vert > C_2 \vert x\vert$ implies that $f_x(y) \ge \Vert Tx\Vert_V$, i.e. $$\inf_{y\in N(T)} f_x(y) =\inf_{y\in K_x} f_x(y)$$ where $K_x=\{y\in N(T)\vert \vert y \vert\le C_2 \vert x\vert\}$. Further note that since $N(T)$ is finite dimensional, all norms are equivalent on $N(T)$ and there exists a constant $C_3$ with $$\Vert y \Vert_V \le C_3 \vert y \vert\quad (y\in N(T)).$$ Consequently $$\Vert x \Vert_V \le \Vert x-y\Vert_V + \Vert y \Vert _V\le f_x(y) + C_3\vert y\vert.$$ Take the infimum over $K_x$, then $$\Vert x \Vert_V \le \inf_{y\in K_x} f_x(y) + C_2C_3\vert x\vert \le C_1 \Vert Tx\Vert_w + C_2C_3\vert x\vert$$

• I think that you left out – Umberto P. May 31 '18 at 14:46
• I accidentally posted to quickly, give me some time to finish the answer. – Jan Bohr May 31 '18 at 14:46