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We define $C_c(\mathbb{R})$ as $\{f: \mathbb{R} → \mathbb{C}: f \;\text{vanishes outside some bounded interval } [a, b]\}$. We want to show that this set is a linear space.

Let $f, g \in C_c(\mathbb{R})$, and let $h = f + g$ be the sum of the two functions. We need to show that $h$ also belongs to $C_c(\mathbb{R})$. From the definition of $C_c(\mathbb{R})$, we know that both $f$ and $g$ vanish outside some bounded interval $[a, b]$. Thus, $f$ and $g$ are zero except within this interval. To show that $h$ vanishes outside a bounded interval as well, let's choose a bounded interval $[c, d]$ that contains the intervals $[a, b]$ which make $f$ and $g$ vanish, respectively. Within this interval, both $f$ and $g$ are zero, and therefore $h = f + g$ will also be zero. Outside this interval, both $f$ and $g$ may or may not be zero. However, this doesn't change the fact that $h$ is zero outside a bounded interval $[c, d]$. Therefore, $h = f + g$ belongs to $C_c(\mathbb{R})$, satisfying the additive property.

Let $f \in C_c(\mathbb{R})$ and let a be a scalar. We need to show that the product $af$ also belongs to $C_c(\mathbb{R})$. From the definition of $C_c(\mathbb{R})$, we know that $f$ vanishes outside a bounded interval $[a, b]$. Thus, $f$ is zero except within this interval. To show that af vanishes outside a bounded interval as well, let's choose the same bounded interval $[a, b]$. Within this interval, $af$ will be equal to the scalar multiplication $a * f(x)$ for every $x$. If $f$ is zero within this interval, then $af$ will also be zero. Outside this interval, both $f$ and $af$ may or may not be zero. However, this doesn't change the fact that $af$ is zero outside the bounded interval $[a, b]$. Therefore, $af$ belongs to $C_c(\mathbb{R})$, satisfying the scalar multiplication property.

By satisfying both the closure under addition and closure under scalar multiplication, we have shown that $C_c(\mathbb{R})$ is a linear space.

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  • $\begingroup$ For some basic information about writing mathematics at this site see, e.g., here, here, here and here. $\endgroup$ Commented Jun 26, 2023 at 13:58
  • $\begingroup$ Welcome to MSE! Make sure to format your math with mathJax for the best response. It is also required that the body of the post can be understood without the title. I edited these for you. It would also help to end your post with a question or statement clarifying exactly what kind of help you'd like. $\endgroup$
    – Daniel
    Commented Jun 26, 2023 at 14:04
  • $\begingroup$ what is your question? For your proof, you should show that $[c,d]$ exists... $\endgroup$ Commented Jun 26, 2023 at 14:15

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First of all, I think you should state more clearly that you are asking if you proof is correct (that is at least what I understand, because of the "solution-verification" tag).

About the proof itself, the general idea is good, but in my opinion, it lacks a bit of rigor and has some typos. I'll try to write the same proof, but more rigorous, and correct the typos.

First step: if $f,g\in C_c(\mathbb{R})$, then $f+g\in C_c(\mathbb{R})$. By definition, there exists one bounded interval $[a_1,b_1]$ such that $f\equiv0$ in $[a_1,b_1]^c$, and the same for $g$, lets call it $[a_2,b_2]$. Then, we can clearly see that $f+g\equiv 0$ in the open set $[c,d]^c$, where $c=min(a_1,a_2)$ and $d=max(b_1,b_2)$, and we conclude $f+g\in C_c(\mathbb{R})$.

(Here what I did is to give an explicit definition of the bounded interval where $f+g$ doesn't vanish, so there is no doubt whether it exist or not, and which clearly illustrates why the function vanishes outside the interval: Also, you made a typo, because you said "let's choose a bounded interval [c,d] that contains the intervals [a,b] which make f and g vanish", when you have precisely the contrary, that there is an interval such that $f$ vanishes on the complement, and same for $g$)

Second step: this one is a bit easier. Let $a\in\mathbb{R}$ a scalar. Again, we know that there is bounded interval $[a_1,b_1]$ such that $f$ vanishes outside it. Therefore, as $(af)(x)=a\cdot f(x)$ by definition, $af\equiv 0$ in $[a_1,b_1]^c$, and therefore $af\in C_c(\mathbb{R})$.

(In this part, the idea was better structured as a formal proof, but you again made the typo of saying that the function $f$ vanishes inside the interval, and not outside. I quote: "Outside this interval, both $f$ and $af$ may or may not be zero". Outside this interval is precisely where you know for sure $f$ will be $0$, and that is what allows you to see that $af$ is also $0$ everywhere outside the interval.)

So we proved that $C_c(\mathbb{R})$ is closed under addition and product by an scalar. But that alone doesn't proof that it is a linear space. You would further need to proof the associativity, conmutativity, existence of identity and inverse of the addition, and the associativity and existence of identity for the product by scalars, and the distributive property w.r.t. addition and product scalar.

However, even though is not very difficult to prove those, you can skip those by noticing that every $f\in C_c(\mathbb{R})$ is also in $C(\mathbb{R})$, the sets of continuous functions with real argument, which we know is a linear space. Then the fact that our set is a subset of a linear space closed under addition and product by scalars it's just the definition of linear subspace, and therefore, $C_c(\mathbb{R})$ inherits the linear space properties from $C(\mathbb{R})$, and we are done.

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