# Continuous inverse of an injective linear function

$$X,Y,Z$$ are Banach spaces, $$A:X\to Y$$ and $$B:X\to Z$$ are continous linear injective functions and $$B$$ is also compact. Moreover, there exists $$C>0$$ such that $$\Vert x\Vert\leqslant C􏰎\Vert Ax\Vert+C􏰎\Vert Bx\Vert$$. To show is there exists some constnat $$D$$ such that $$\Vert x\Vert\leqslant D\Vert Ax\Vert$$.

I would like to apply the continuous inverse theorem or open mapping theorem. But I need the image of $$A$$ in $$Y$$ is closed. However, I am not able to show it.

If this is false then there exist $$x_n$$'s such that $$\|x_n\|>n\|Px_n\|$$. Let $$y_n=\frac {x_n} {\|x_n\|}$$. Since $$y_n$$ is bounded there is a subsequence of $$Q(y_n)$$ which is convergent. Let $$Q(y_{n_i}) \to z$$. Then $$\|y_{n_i}-y_{n_j}\| \leq C\|P(y_{n_i}-y_{n_j})\|+C\|Q(y_{n_i}-y_{n_j})\| \to 0$$ since $$Py_{n_i} \to 0$$. Let $$y =\lim y_{n_i}$$. Then $$\|y\|=1$$ but $$Py=0$$ a contradiction to injectivity.
• $y_n$ is $x_n$ normalized? – CO2 Aug 1 at 12:05