# Null Space of Bounded Linear Functionals

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.

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

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

thus,

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

this clearly describes a line through the origin; likewise if

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

then

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

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

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

then

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

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

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

if

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

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.