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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:

enter image description here

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 $(10)$)

enter image description here

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2 Answers 2

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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 $$f(x) = f(\alpha x_0) = \alpha f(x_0) = \alpha \|x_0\|.$$

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The only constant linear functional on any vector space is 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.

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  • $\begingroup$ yeah my post says the affine functions, your answer is true but can you check the question again I added the reason which I asked this question. $\endgroup$ Jun 9, 2020 at 1:10
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    $\begingroup$ I responded to your edited question, take a look $\endgroup$ Jun 9, 2020 at 1:40

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