Linear independence of two functions, how to solve I have a problem.  I need to show that $\{f,g\}$ is a linearly independent set in the vector space of all functions from $\mathbb{R}^{+}$ to $\mathbb{R}$, where
$$f(x)=x$$
$$g(x)=\frac1{x}$$
First (and least important), is there a standard notation for this vector space?
Second, I didn't know how to answer this question, so I cheated and checked the solution manual.  There, it basically showed that letting $x=1$ first and $x=2$ second and setting $h(x)=af(x)+bg(x)=0$ gives a system of equation.  Solving shows $a=b=0$, and thus linearly independent.  
Not having any exposure really with this particular vector space, why are we able to pick values for $x$?  We don't do that for $\mathbb{R}^n$ or $\mathbb{F}_n[x]$.  I guess I'm just not accustomed to solving problems in this vector space and am unsure how to procede and why this is the method?  Can someone explain to me why this is the correct way?  
 A: The space of all functions from $\mathbb R^+$ to $\mathbb R$ is $\mathbb R^{(\mathbb R^+)}$. If you only want the continuous ones, then it's denoted by $C(\mathbb R^+,\mathbb R)$.
Now, if any two functions $f,g:\mathbb R^+\to\mathbb R$ are linearly dependent, then, by definition, there exists non-zero scalars $a,b$ such that $af+bg=0$. Now the latter is a functional equation, namely it holds for all $x>0$:$af(x)+bg(x)=0$. So, in particular, it holds for the values $x=1$ and $x=2$. But, as it turns out, the only solution to $af(1)+bg(1)=0$ and $af(2)+bg(2)=0$ is $a=b=0$, thus you could not have the non-zero $a,b$ above. This shows that assuming $f,g$ were dependent leads to a contradiction and thus they are independent. 
A: Showing linear independence amounts to showing that
$$
\lambda_1f + \lambda_2g = 0 \implies \lambda_1 = \lambda_2 = 0
$$
Equality for functions (usually) means pointwise equality, hence you can assume
$$
\lambda_1f(x) + \lambda_2g(x) = 0
$$
for all $x \in \mathbb{R}_+$. This means
$$
\lambda_1x + \lambda_2\frac{1}{x} = 0
$$
which is equivalent to
$$
\lambda_1x^2 = -\lambda_2
$$
Assuming that $\lambda_1$ is not zero we obtain
$$
x^2 = - \frac{\lambda_2}{\lambda_1}
$$
for any $x \in \mathbb{R}_+$. Which is obviously a contradiction (take for example $x=1$ and $x=2$ or any other two distinct values - nothing so special about these two choices for $x$ - they just lead to a shorter argument). Hence $\lambda_1 = 0$ and therefore by the previous equation also $\lambda_2 = 0$.
Not sure if there is any standard notation for this vector space, usually you are interested in functions spaces with at least a little bit more structure.
A: The standard notation for the vector space of all functions $\Bbb R^+\to\Bbb R$ is $\Bbb R^{\Bbb R^+}$. This is sort of awkward and I've also seen the notation $\mathcal F(0,\infty)$ used. As long as you define your notation properly I don't think notation matters much here.
As for showing that $f(x)$ and $g(x)$ are linearly independent in $\mathcal F(0,\infty)$, we can use the definition. Seeking a contradiction, suppose that there exist scalars $a$ and $b$ not both zero such that
$$
af(x)+bg(x)=0\tag{1}
$$
for all $x>0$. Then 
$$
ax=-\frac{b}{x}\tag{2}
$$
for all $x>0$. In particular, plugging $x=1$ and $x=2$ into (2) gives
$$
a=-b\tag{3}
$$
and
$$
2a=-\frac{b}{2}\tag{4}
$$
Plugging (3) into (4) gives
$$
2a=\frac{a}{2}
$$
which implies $a=0$. But then (3) implies $b=0$ too, a contradiction. Hence $f(x)$ and $g(x)$ are linearly independent in $\mathcal F(0,\infty)$.
The key point here is that the equation (1) holds for all $x>0$ which is what allows us to arrive at our desired contradiction. The reason (1) is assumed to hold for all $x>0$ is because the elements of $\mathcal F(0,\infty)$ are functions which are equal if and only if they are equal at every point of $\Bbb R^+$.
A: The set $\{f,g\}$ is linearly independent if 
$$af+bg=0\Rightarrow a=b=0$$
which we can write it also in this form
$$af(x)+bg(x)=0,\quad\forall x\in D_f\cap D_g\Rightarrow a=b=0$$
so the above equality is valid for all $x$ in the domain of $f$ and $g$ and then we can take some particular values of $x$ to determine a necessary condition of $a$ and $b$.
Notice also that
$$af(x)+bg(x)=0\iff ax^2+bx=0,\quad\forall x\ne0$$
and since the set of polynomials $\{x,x^2\}$ is  linearly independent in $\Bbb R[x]$ we deduce that $a=b=0$  so we find the same result.
A: Your statement is that for two constants $a$ and $b$ the function $h(x) = af(x)+bg(x)$ is zero for all values of $x \in \mathbb{R}^+$.
One way to prove independence in cases like this it is taking the derivative of $h(x)$ and assuming $h(x) \equiv 0$. This leads to
$$a - \frac{b}{x^2} = 0$$ for all values of $x$ what is clearly false.
