$\operatorname{Ker}(B) \subset \operatorname{Ker(A)}$ if and only if ${\left\| {Ax} \right\|_X} \leqslant \alpha {\left\| {Bx} \right\|_X}$

Question

Is this statement is correct: Let $X$ be a Banach space, and let $A$ and $B$ two continuous operators on $X$ , do we have for some constant $\alpha$ the following $${\left\| {Ax} \right\|_X} \leqslant \alpha {\left\| {Bx} \right\|_X} \Leftrightarrow \operatorname{Ker}(B) \subset \operatorname{Ker}(A)$$. Thank you.

Answer 1

In finite dimensions, it should be true. Let us denote by $R$ the range of $B$ and by $S$ the range of $A$. Then, you can define the linear operator $$A \, B^{-1} : R \to S.$$ Due to the assumptions on the kernels, this operator is well defined and linear. Thus, it is continuous, i.e., there is $\alpha \ge 0$ with $$ \|A \, B^{-1} y \| \le \alpha \, \|y\| \qquad\forall y \in R.$$ Now, setting $y = B \, x$ yields the claim.

Answer 2

An example in the infinite-dimensional case can be constructed as follows: Let $A=Id$ and $B:X\to X$ compact and injective. Thus $\ker A=\ker B$. If there would be $\alpha>0$ such that $\|Ax\|=\|x\|\le \alpha \|Bx\|$ then the image of $B$ would be closed, thus $B$ would be bijective and continuously invertible, which is a contradiction.

An easy example is given by $X=l^2$ and $$ Bx = (x_1,x_2/2,x_3/3,\dots,x_n/n,\dots). $$ Here you could prove by hand that no such $\alpha$ exists (without recurring to compactness).