Bounding linear functional

Question

Bounding a linear functional in $L_2[0, 1]$ I came across this question today. I understand the accepted answer integrates the $ay$ part to get $l(f)$ when $a$ is not $0$. However, I'm stucked when $a=0$, its not that obvious to me that the method used in accepted answer still works since the solution of $y'=f$ is just the integral of $f$ and I don't know how to use Cauchy-Schwartz to get the bounded part. So can anyone help me with the case when $a=0$? Any help is appreciated.

Answer 1

If $a=0$, then $\phi(t) = \int_0^tf(s)\,ds$ and thus (using Fubini) $$ l(f) = \int_0^1\int_0^tf(s)\,ds\,dt = \int_0^1f(s)\int_s^1\,dt\,ds = \int_0^1(1-s)f(s)\,ds. $$ Hence, $$ |l(f)|\le \int_0^1(1-s)|f(s)|\,ds\le\left(\int_0^1(1-s)^2\,ds\right)^{1/2}\|f\|_2 = \frac 1{\sqrt 3}\|f\|_2. $$