Question: I must be missing something very simple but may I quickly ask how to show that the linear operator $f: \ell^1 \rightarrow (c\_0)'$, $f\_y(x)= \sum \limits\_{i=1}^\infty x\_i\*y\_i\ $ from [linear operator from $\ell^1$ to $(c\_0)'$](https://math.stackexchange.com/questions/2308372/linear-operator-from-ell1-to-c-0?noredirect=1&lq=1) is bounded?

Answer: 
First differential equation 3 to get
$\frac{dy\left(t\right)}{dt}=w\cos\left(wt\right)\cos\phi-w\sin\left(wt\right)\sin\phi$. Then subtract $y\left(t\right)w\mathrm{ctg}\phi$ from both sides of the equation and simplify. You'll get a cancelation of the $w\cos\left(wt\right)\cos\phi$ term, and finally use the trig identity $\sin^{2}\phi+\cos^{2}\phi=1$ to simplify the remaining 2 terms.
So, we have
$$
y(t) = \sin(\omega t + \phi) = \sin(\omega t)\cos(\phi) + \cos(\omega t)\sin(\phi).
$$
Differentiating this with respect to $t$ gives
$$
y'(t) = \omega\cos(\omega t + \phi) = \omega\cos(\omega t)\cos(\phi)-\omega\sin(\omega t)\sin(\phi).
$$
At this point, the author wants to eliminate $\cos(\omega t)$ from the equations, as it has no simple expression in terms of $y$ and $x$. Some quick linear algebra gives
$$
\sin(\phi)y'(t) - \omega\cos(\phi)y(t) = -\omega\sin(\omega t)\left[\cos(\phi)^2+ \sin(\phi)^2\right] = -\omega \sin(\omega t).
$$
Now all that remains is to substitute $\sin(\omega t) = x(t)$ and divide through by $\sin(\phi)$ to get
$$
y'(t) - \omega\cot(\phi)y(t) = -\omega\csc(\phi)x(t).
$$