# Constant functionals on a set generated by one elements are linear?

In functional analysis we say that if $$X$$ is a normed space and $$f\in X'$$(dual of $$X$$).(Say real normed space.)

f is $$f:X\to \mathbb R$$ to satisfy $$f(x+y)=f(x)+f(y),\\f(\alpha x)=\alpha x \\ \forall x,y \in X, \forall \alpha \in \mathbb R$$

But If I want to consider $$Z=span(\{x_0\})\subset X$$ and define $$f$$ as

$$f(x)=1,\quad \forall x\in Z$$

How can I formally show that $$f$$ is linear.

The intuition is clear because $$f$$ is constant functional on $$Z$$ so it is linear but formally $$1=f(x+y)\neq f(x)+f(y)=1+1=2$$

So what is the problem?

I want such a thing to show that functional defined on $$Z$$ is bounded.

My problem arises from a solution from functional analysis problem below: And another example is that using "linear functional" in a sense that I want to understand is from Kreyszig Functional Analysis Book. (Look at the line described by $$(10)$$) The proposed constant function is not linear. While the graph of $$f$$ forms a line, it doesn't pass through the origin, making it an affine function, not a linear function.
In the given solution, $$f$$ is not being defined to be constantly $$1$$, but is being defined to be $$1$$ at the single point $$x - y$$. Since $$x - y$$ is a basis for $$\operatorname{span}(x - y)$$, this is sufficient to define a linear map. In particular, the linear map takes an arbitrary point $$\alpha(x - y) \in \operatorname{span}(x - y)$$, and maps it to $$\alpha$$. Really though, you don't need to understand what happens further along the line $$\operatorname{span}(x - y)$$, so long as $$f$$ maps $$x - y$$ to something non-zero.
Your third question has much the same answer. Don't forget that the $$x$$ in $$(10)$$ takes the form $$x = \alpha x_0$$ (which is the form that all elements of $$\operatorname{span}(x_0)$$ take). Then, using linearity, and the fact that $$f(x_0)$$ is defined to be $$\|x_0\|$$, we get $$f(x) = f(\alpha x_0) = \alpha f(x_0) = \alpha \|x_0\|.$$
Addendum after edit: so, the idea is that if you want to define a linear functional on a line, you can choose a non-zero point (which will be a basis) and map it to any real number you want. In this case, you fixed $$x_0$$ and want to map out it to $$\|x_0\|$$. So what do you map $$-7x_0$$ to? You map it to $$-7\|x_0\|$$. On this line, you have operator norm 1. Now Hahn-Banach says you can extend this guy to a linear functional on the whole space.