Bounding a linear functional in $L_2[0, 1]$ 
For each f in $L_2[0, 1]$ let $\phi(t)$ be the solution of $y' + ay = f$ that satisfies $\phi(0) = 0$, where a is a constant. Define $l: L_2[0,1] \to \mathbb{C}$ by $l(f) = \int_0^1 \phi(t) dt.$

I'm able to use just the property $\phi' + a\phi = f$ to show this functional is linear, and I'm at least reasonably sure that in general, $\phi$ should look like $e^{-at}\int_0^t e^{as} f(s)ds$, so that ultimately $l(f) = \int_0^1 f(t) (1-e^{at})dt$ after some messing with the double integral that results from substituting it into the formula for l(f). There are two problems, though: First, I'm not sure if what I think $\phi$ is always exists, since this is $L_2$, so possibly (at least in theory) the integral $\int_0^t e^{as} f(s)ds$ doesn't converge in some non-null set. Second, I can't seem to find any way to actually bound $l(f)$ in terms of $||f|| = \sqrt{\int_0^1 |f(t)|^2 dt}$. It would be best if there is a way to show that l(f) is bounded without using the value I think $\phi$ has, since it's not (or at least, I don't think it is) clear that $\int_0^1 \phi(t) = \int_0^1 f(t) (1-e^{at})dt$ exists without already knowing that l(f) is bounded, but both of my problems are solved if I can get that bound.
I tried looking at $||f||^2$ in terms of $f = \phi' + a\phi$, but this didn't pan out, since, for example, $|\int_0^1 (\phi'(t) + a\phi(t))dt |^2$ might not have any meaningful inequality to relate it to l(f).
 A: Suppose that $a\ne0$, and consider first the case of continuous functions $f\in L^2([0,1])$, in this case
$(e^{at}\phi(t))'= e^{at}f(t)$. So, since $\phi(0)=0$,
$$e^a\phi(1)=\int_0^1e^{at}f(t)dt\tag{1}$$
Now, integrating $\phi'+a\phi=f$ on $[0,1]$, we get
$$\phi(1)+a\,l(f)=\int_0^1f(t)dt\tag{2}$$
Combining $(1)$ and $(2)$ we obtain
$$
l(f)=\frac{1}{a}\int_0^1(1-e^{a(t-1)})f(t)dt\tag{3}
$$
By Cauchy-Schwarz inequality we get
$$
\left\vert l(f)\right\vert^2\leq\frac{1}{a^2}\int_0^1\left(1-e^{a(t-1)}\right)^2dt\cdot
\int_0^1\left\vert f(t)\right\vert^2dt
$$
Or equivalently
$$|l(f)|\leq M_a \left\Vert f\right\Vert_2
\quad\hbox{with $M_a=\sqrt{
\dfrac{2 a+4 e^{-a}-e^{-2 a}-3}{2 a^3}}$}$$
This proves that $l$, defined by $(3)$, has a unique continuous extension to $L^2([0,1])$.
 and that $\Vert l\Vert\leq M_a $. In fact we have equality since in this inequality, since it is achieved if $f$ is proportional to $t\mapsto \frac{1}{a}(1-e^{a(t-1)})$.
Finally, the case $a=0$ is obtained in the same lines as before, or by continuity, with $M_{0}=1/\sqrt{3}$.
