$X,Y,Z$ 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.
Best Answer
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.