I am attempting to solve the following problem:
Let $V$ be the vector space of all continuous functions $f : R → R$ with point-wise addition and scalar multiplication defined.
(a) Show that
$M_1$ = {$f ∈ V :$ There exists $a, b ∈ R$ with $f(x) = a sin(x + b)$ for all $x ∈ R$}
is a linear subspace of $V$.
(b) Show that
$M_2$ = {$f ∈ V :$ There exists $a ∈ R$ with $f(x) = sin(ax)$ for all $x ∈ R$}
is not a linear subspace of $V$.
As far as I understand, the requirement for being a linear subspace is that the set must be algebraically closed with respect to addition and multiplication by a scalar, that is,
(i) If $f_1,f_2 ∈ V$ then $f_1+f_2 ∈ V$
(ii) If $f ∈ V$ then $λf ∈ V$, where $λ∈R$
However, trying to apply the first condition to prove part (a), I need to find a way to express
$(f_1+f_2)(x)=f_1(x)+f_2(x)=a_1sin(x+b_1)+a_2sin(x+b_2)$
in the form
$asin(x+b)$
so that I can prove that the sum of the two functions belongs to $M_1$ as well.
Further, the second condition (multiplication by a scalar) is valid as $(λf)(x)=λasin(x+b)$ which belongs to $M_1$. It is just the first condition at which I am stuck.
Also, for part (b), I would appreciate hints for proving how the set is not a linear subspace? What would be a way of showing conclusively that the two conditions do not hold?
Best Answer
You have \begin{align} a_1\sin(x+b_1)+a_2\sin(x+b_2)&=a_1\sin x\cos b_1+a_1\cos x\sin b_1+a_2\sin x\cos b_2+a_2\cos x\sin b_2\\ &=(a_1\cos b_1+a_2\cos b_2)\sin x+(a_1\sin b_1+a_2\sin b_2)\cos x. \end{align} So it is enough to show that $\alpha\sin x+\beta\cos x\in M_1$. For this, note that we can always find $t$ with $$ \frac\alpha{\sqrt{\alpha^2+\beta^2}}=\cos t,\ \ \ \frac\beta{\sqrt{\alpha^2+\beta^2}}=\sin t. $$ So \begin{align} \alpha\sin x+\beta\cos x&=\sqrt{\alpha^2+\beta^2}\,(\cos t\sin x+\sin t\cos x)\\ &=\sqrt{\alpha^2+\beta^2}\,\sin(x+t)\in M_1. \end{align} For $M_2$, as Alex noted, all elements $\sin (ax)$ have absolute value $\leq1$, so $5\sin x$ cannot be written as $\sin (ax)$.