Necessary and Sufficient Conditions for a Linear Functional to Be Expressed as a Linearly Independent Combination
Theorem
Let $f, f_{1} , \cdots , f_{n}$ be linear functionals whose domain is $X$.
(a) For $c_{1} , \cdots , c_{n} \in \mathbb{C}$, $\displaystyle f = \sum_{i=1}^{n} c_{i} f_{i}$ $\iff$ $\displaystyle \bigcap_{i=1}^{n} \ker ( f_{i} ) \subset \ker (f)$
(b) $f_{1} , \cdots , f_{n}$ are linearly independent $\iff$ there exist $x_{1} , \cdots , x_{n}$ satisfying $f_{j} (x_{i} ) = \delta_{ij}$.
Here, $\delta_{ij}$ is the Kronecker delta.
Explanation
Considering that the kernel is related to the concept of being homogeneous, one can guess that this is a useful fact for linear homogeneous differential equations. However, from a learner’s point of view, the proof is excessively long, difficult, and complicated, so it is recommended to simply keep the fact itself well in mind.
Proof
(a)
Strategy: $( \implies )$ It can be shown easily using only the definition of the kernel. $( \impliedby )$ We concretely find $c_{1} , \cdots , c_{n}$.
$(\implies )$
Since $\displaystyle x \in \bigcap_{i=1}^{n} \ker ( f_{i} )$,
$$ f_{i} ( x ) = 0 $$
Since $\displaystyle f(x) = \sum_{i=1}^{n} c_{i} f_{i} (x) = 0$,
$$ x \in \ker (f) $$
To summarize,
$$ \bigcap_{i=1}^{n} \ker ( f_{i} ) \subset \ker (f) $$
$( \impliedby )$
Define the proposition $ \displaystyle P(n) : \bigcap_{i=1}^{n} \ker ( f_{i} ) \subset \ker (f) \implies f = \sum_{i=1}^{n} c_{i} f_{i}$ and use mathematical induction.
Since the case $f = 0$ is trivial, assume $f \ne 0$.
Part 1. $n=1$
Since $\ker (f_{1} ) \subset \ker (f)$,
$$ f_{1} \ne 0 $$
and there exists $x_{1} \in X$ satisfying
$$ f_{1} (x_{1} ) = 1 $$
If $x \in X$, then $x - f_{1} (x) x_{1} \in X$, and applying $f_{1}$ to it gives
$$ f_{1} ( x - f_{1} (x) x_{1} ) = f_{1} ( x ) - f_{1} (x) f_{1} ( x_{1} ) = 0 $$
That is, $x - f_{1} (x) x_{1} \in \ker (f_{1} ) \subset \ker (f)$, and applying $f$ to it gives
$$ 0 = f ( x - f_{1} (x) x_{1} ) = f(x) - f_{1} (x) f_{1} (x_{1}) $$
To summarize, for $f_{1} (x_{1}) \in \mathbb{C}$, $f$ can be expressed as $f(x) = f_{1} (x_{1}) f_{1} (X)$.
Part 2. $n=N-1$
Assume that $P(N-1)$ holds.
Part 3. $n=N$
Case 1. When $f_{1} , \cdots , f_{N}$ are not linearly independent
Since $f_{1} , \cdots , f_{N}$ are not linearly independent, there exist some $t_{i} \in \mathbb{C}$ satisfying
$$ t_{1} f_{1} + \cdots + t_{N} f_{N} = 0 $$
$$ t_{i_{0}} \ne 0 $$
Since $f_{i_{0}}$ is expressed as
$$ \displaystyle f_{i_{0}} = {{1} \over { t_{i_{0}} }} \left( \sum_{i \ne i_{0}} t_{i} f_{i} \right) =\sum_{i \ne i_{0}} \left( {{t_{i} } \over { t_{i_{0}} }} \right) f_{i} $$
we have
$$ \bigcap_{i \ne i_{0} } \ker ( f_{i} ) \subset \ker (f_{ i_{0}} ) $$
Since $\displaystyle \bigcap_{i \ne i_{0} } \ker ( f_{i} )$ is contained in $\displaystyle \ker (f_{ i_{0}} )$,
$$ \bigcap_{i \ne i_{0} } \ker ( f_{i} ) = \left[ \bigcap_{i \ne i_{0} } \ker ( f_{i} ) \right] \cap \ker (f_{i_{0}} ) = \bigcap_{i=1}^{ N } \ker ( f_{i} ) \subset \ker (f) $$
But since we assumed in Part 2. that $P(N-1)$ holds, there exist $c_{1} , \cdots , c_{N-1} \in \mathbb{C}$ satisfying $\displaystyle f = \sum_{ i \ne i_{0}} c_{i} f_{i} + 0 f_{i_{0}}$.
Case 2. When $f_{1} , \cdots , f_{N}$ are linearly independent
If we suppose $\displaystyle \bigcap_{ k \ne i } \ker ( f_{k} ) \subset \ker (f_{i} )$ for $1 \le k \le N$, then since we assumed in Part 2. that $P(N-1)$ holds, $f_{i}$ would be expressed as $\displaystyle f_{i} = \sum_{ k \ne i } \lambda_{k} f_{k}$ for some $\lambda_{1} , \cdots , \lambda_{N} \in \mathbb{C}$, and $f_{1} , \cdots , f_{N}$ would not be linearly independent; hence it must be that $\displaystyle \bigcap_{ k \ne i } \ker ( f_{k} ) \not\subset \ker (f_{i} )$. Then there exist $y_{1} , \cdots , y_{N} \in X$ satisfying $$ \displaystyle y_{i} \in \left[ \bigcap_{ k \ne i } \ker ( f_{k} ) \right] \setminus \ker (f_{i} ) $$
$$ y_{i} \in \ker (f_{i} ) $$
For these, defining $\displaystyle x_{i} := {{ y_{i}} \over {f_{i} ( y_{i} ) }}$ gives
$$ \begin{cases} \displaystyle f_{j} (x_{i} ) = {{ f_{j} (y_{i}) } \over { f_{i} (y_{i} ) }} = 0 \\ \displaystyle f_{i} (x_{i} ) = {{ f_{i} (y_{i}) } \over { f_{i} (y_{i} ) }} = 1 \end{cases} \implies f_{j} ( x_{ i} ) = \delta_{ij} = \begin{cases} 0 & , i \ne j \\ 1 & , i = j \end{cases} $$
Now, applying $f_{i}$ to $\displaystyle x - \sum_{i=1}^{N} f_{j} (x) x_{j}$, defined for arbitrary $x \in X$, gives for all $i = 1 , \cdots , N$
$$ \begin{align*} f_{i } \left( x - \sum_{j=1}^{N} f_{j} (x) x_{j} \right) =& f_{j} (x) - \sum_{i=1}^{N} f_{j} (x) f_{i} ( x_{j} ) \\ =& f_{i} (x) - \sum_{j=1}^{N} f_{j} (x) \delta_{ij} \\ =& f_{i} (x) - f_{i} (x) \\ =& 0 \end{align*} $$
Expressing this as a set inclusion,
$$ \left( x - \sum_{j=1}^{N} f_{j} (x) x_{j} \right) \in \bigcap_{i=1}^{N} \ker ( f_{i} ) \subset \ker (f) $$
By the definition of the kernel, applying $f$ to $\displaystyle x - \sum_{j=1}^{N} f_{j} (x) x_{j}$ gives
$$ f \left( x - \sum_{j=1}^{N} f_{j} (x) x_{j} \right) = f(x) - \sum_{j=1}^{N} f_{j} (x) f ( x_{j} ) = 0 $$
$$ \implies f(x) =\sum_{j=1}^{N} f_{j} (x) f ( x_{j} ) = \left[ \sum_{j=1}^{N} f_{j} f ( x_{j} ) \right] (x) $$
$$ \implies f = \sum_{j=1}^{N} f ( x_{j} ) f_{j} $$
That is, $f$ is expressed as a linear combination of $f_{1} , \cdots , f_{N}$ with concrete $f ( x_{1} ) , \cdots , f ( x_{N} ) \in \mathbb{C}$.
Therefore, whether $f_{1} , \cdots , f_{N}$ are linearly independent or not, the proposition $P(n)$ holds for all $n \in \mathbb{N}$.
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(b)
Strategy: It is essentially a corollary of (a).
$(\implies)$
Indeed, in Proof (a)-$(\impliedby)$-Part 3.-Case 2., when $f_{1} , \cdots , f_{n}$ were linearly independent, there existed $\displaystyle x_{i} := {{ y_{i}} \over {f_{i} ( y_{i} ) }}$ satisfying $f_{j} (x_{i} ) = \delta_{ij}$.
$(\impliedby)$
When there exist $x_{1} , \cdots , x_{n}$ satisfying $f_{j} (x_{i} ) = \delta_{ij}$, consider $c_{1} f_{1} (x) + \cdots + c_{n} f_{n} (x) = 0$. Substituting $x_{1}$ gives
$$ c_{1} f_{1} (x_{1} ) + 0 + \cdots + 0 = c_{1} = 0 $$
Likewise, substituting $x_{i}$ gives
$$ 0 + \cdots + 0 + c_{i} f_{i} + 0 + \cdots + 0 = c_{i} = 0 $$
Therefore, the only case satisfying $c_{1} f_{1} (x) + \cdots + c_{n} f_{n} (x) = 0$ is $c_{1} = \cdots = c_{n} = 0$, and $f_{1} , \cdots , f_{n}$ are linearly independent.
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