How to solve the following:
Let $X$ and $Y$ be normed vector spaces and let $L_1 : X → Y$ and $L_2 : X → R$ be linear operators such that $KerL_1 ⊂ KerL_2$. Prove that exists linear operator $λ : Y → R$, such that $L_2 = λL_1$.
Any hint or help is welcome. Thanks in advance.
This is a standard fact for linear spaces. Let $B=\{e_i:i\in I\}$ be any basis of the space $\operatorname{Im} L_1$. We can extend it to a basis $B’=\{e_i:i\in I’\}$ of the space $Y$. Define a linear operator $\lambda: Y\to R$ as follows. First define $\lambda(e_i)$ for each $e_i\in B’$. If $e_i\in B$ then pick any vector $f_i\in L_1^{-1}(e_i)$ and put $\lambda(e_i)=L_2(f_i)$. If $e_i\in B’\setminus B$ then put $\lambda(e_i)=0$. Now let $y\in Y$ be an arbitrary element. Then there exists a finite subset $J$ of the index set $I$ and a family $\{c_i:i\in J\}$ of scalars such that $y=\sum_{i\in J} c_ie_i$. Put $\lambda y=\sum_{i\in J} c_i \lambda e_i$. Since $B’$ is a basis then (it is easy to check that) $\lambda$ is a linear operator. Let $x\in X$ be an arbitrary element and $L_1x=\sum_{i\in J} c_ie_i$ for a family $\{c_i:i\in J\}$ of scalars. Then $$\lambda L_1x -L_2x=\sum_{i\in J} c_i L_2(f_i)-L_2x=L_2\left(\sum_{i\in J} c_if_i-x\right).$$ Since $\operatorname{Ker} L_1\subset\operatorname{Ker} L_2$ to show that $\lambda L_1x -L_2x=0$ it suffices to remark that $$L_1\left(\sum_{i\in J} c_if_i-x\right)= \sum_{i\in J} c_i e_i-L_1x=0.$$