Show that the mapping $I : C^0(\Bbb R) → C^1(\Bbb R)$ defined by $I(f)(x) =\int_2^x f(t)dt$ is a linear transformation.
Am I just proving that $I(v+w)= I(v)+C(w)$ and $I(cv)=cI(v)$?
And would that just involve replacing the x in the integral with $u$ and $v$?
On
As you said,all you have to do is to prove that $I(f + g) = I(f) + I(g)$ and that $I(c*f) = c*I(f)$ for any real number c. For the first,notice that: $I(f+g)$ = integral of $f+g$ = integral of $f$ + integral of ($g$) (from 2 to $x$ ) For the second, $I(c*f)$ = integral of $c*f$ = c* integral of ($f$) Then the map given is linear.
No. Here $u$ and $v$ are elements of $C^0(\Bbb R)$, not $\Bbb R$.
Show that $$\int_2^x (cu)(t)dt=c\int_2^x u(t)dt$$ and $$\int_2^x (u+v)(t)dt=\int_2^x u(t)dt+\int_2^x v(t)dt.$$