# (Conway's Functional Analysis) Why can we assume that $\bigcap_{j\neq k}ker(f_j) \neq \bigcap_{j=1}^n ker(f_j).$

In Conway's A Course in Functional Analysis, the author stated the following Proposition $1.4$ at page $377.$

Let $X$ be a vector space over $\mathbb{R}.$ Given $f,f_1,f_2,...,f_n$ are linear functionals on $X.$ If $$\bigcap_{j=1}^n ker(f_j) \subseteq ker(f),$$ then there exist scalars $a_1,...,a_n$ such that $$f(x) = \sum_{j=1}^n a_j f_j.$$

The proof starts as follows:

It may be assumed that without loss of generality that for $1\leq k\leq n,$ $$\bigcap_{j\neq k}ker(f_j) \neq \bigcap_{j=1}^n ker(f_j).$$

Question: Why can we assume above? In general, we have $$\bigcap_{j\neq k} ker(f_j) \supseteq \bigcap_{j=1}^n ker(f_j).$$ So in the proof, we can safely assume that the inclusion is strict. If the inclusion is not strict, meaning the two intersections are equal, then kernel of $f_j$ is a subset of kernel of $f_i$ for some $i.$

• If there is an equality, you can just forget about $f_k$ (because the assumption still holds for $f,f_1,\dots,f_{k-1},f_{k+1},\dots,f_n$) and obtain a linear combination where $f_k$ does not appear. Jan 25, 2018 at 9:48

I am sure you know the result which says if $f,g$ are linear functional with $ker(g)\subseteq ker(f)$ then $f$ is a constant multiple of $g$.
Suppose $f,g,h$ are such that $ker(g)\cap ker(h)\subset ker(f)$. Suppose we have $ker(g)\cap ker(h)=ker(g)$ then $ker(g)\subset ker(f)$ then by above fact we have $f=a g=ag+0h$. Similarly in the case when $ker(g)\cap ker(h)=ker(h)$.
You can generalise this for more than$3$ functions.