Assume $f$ is a linear bounded functional from $\mathbb{R}^2\rightarrow\mathbb{R}$. I understand that the null space of $f$, denoted $\mathcal{N}(f)$, is a subspace of $\mathbb{R}^2$. However, I am having a difficult time figuring out why $\mathcal{N}$ must be a line through origin.

I have looked at Griffel's Functional Analysis Chapter 7 and also Zeidler's book but could not find an easy way to justify this.

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    $\begingroup$ Can it not be also be the whole space? Unless you demand from it to not be identically zero? $\endgroup$ – Keen-ameteur Feb 27 at 17:00
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    $\begingroup$ It also seems that the 'bounded' part is a given, since this is a finite dimensional linear map. $\endgroup$ – Keen-ameteur Feb 27 at 17:03

Technically it doesn't.

If $f=0$ then $\mathcal{N}(f) = \mathbb{R}^2$. But otherwise, the image of $f$ is of dimension at least $1$. Moreover since the image of $f$ is a subspace of $\mathbb{R}$ the dimension of the image is also at most $1$. Thus, by the rank-nullity theorem we have that $\mathcal{N}(f)$ is a vector space of dimension $2-1=1$.

Any vector space of dimension $1$ on $\mathbb{R}^2$ takes the form $\text{span}(v)$ where $v\not =0$ is some vector of $\mathbb{R}^2$. Hence, is a line which passes through the origin in the direction of $v$.

One last thing. The word "bounded" is irrelevant here. All linear functionals from $\mathbb{R}^2$ to $\mathbb{R}$ are bounded.


A plain, old-fashioned, vanilla-flavored, very basic demonstration which doesn't even need the rank-nullity theorem, though that is surely a good theorem:

Let us assume that

$f \ne 0; \tag 1$

for otherwise, we clearly have

$\mathcal N(f) = \Bbb R^2, \tag 2$

the trivial case. Now for any such linear $f$, trivial or not, $(x, y) \in \mathcal N(f)$ when

$f(x, y) = f(x(1, 0) + y(0, 1)) = xf(1, 0) + yf(0, 1) = 0; \tag 3$

now if $f \ne 0$, at least one of

$f(1, 0), f(0, 1) \ne 0; \tag 4$

suppose then that

$f(1, 0) \ne 0; \tag 5$


$x = -\dfrac{f(0, 1)}{f(1, 0)} y; \tag 6$

this clearly describes a line through the origin; likewise if

$f(0, 1) \ne 0 \tag 7$


$y = -\dfrac{f(1, 0)}{f(0, 1)} x;, \tag 8$

also the equation of a line through $(0, 0)$.

Note Added in Edit, Wednesday 27 February 2019 10:16 AM PST: The above easily generalizes to show that for linear

$f:\Bbb R^n \to \Bbb R, \tag 9$

$\mathcal N(f)$ is a hyperplane containing $0 \in \Bbb R^n$; for if we adopt the standard basis

$\mathbf e_i \in \Bbb R^n, \; 1 \le i \le n, \tag{10}$


$f \left ( \displaystyle \sum_1^n x_i \mathbf e_i \right ) = \displaystyle \sum_1^n x_i f(\mathbf e_i); \tag{11}$

if $f \ne 0$ then at least one of

$f(\mathbf e_i) \ne 0; \tag{12}$


$f(\mathbf e_j) \ne 0, \tag{13}$

we have via (11),

$f \left ( \displaystyle \sum_1^n x_i \mathbf e_i \right ) = 0 \Longrightarrow x_j = -\displaystyle \sum_{j \ne i = 1}^n \dfrac{f(\mathbf e_i)}{f(\mathbf e_j)} x_i, \tag{14}$

the equation of a hyperplane containing the origin. End of Note.


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