# Textbook problem on dual space of $L_2[0,1]$

Define the linear functional $f$ on $L_2[0,1]$ by $f(x)=\int_0^1a(t)\int_0^tb(s)x(s)dsdt$ where $a,b\in L_2[0,1]$. Show that $f$ is bounded on $L_2[0,1]$ and find an element $y\in L_2[0,1]$ such that $f(x)=(x|y)$.

Note: $(x|y)$ denotes the inner product of $x$ and $y$.

Here is my current progress. First, note that $\|\int_0^tb(s)x(s)ds\|\leq\|b\|\|x\|$, since $$\int_0^1\left(\int_0^tb(s)x(s)ds\right)^2dt\leq\int_0^1\left(\int_0^t|b(s)x(s)|ds\right)^2dt~~~\mbox{(since \forall s~~b(s)x(s)\leq|b(s)x(s)|)}$$ $$\leq\int_0^1\left(\int_0^1|b(s)x(s)|ds\right)^2dt~~~\mbox{(since \forall s~|b(s)x(s)|\geq0)}$$ $$\leq\int_0^1(\|b\|\|x\|)^2dt~~~\mbox{(Hölder inequality)}$$ $$=\|b\|^2\|x\|^2.$$ Thus, $$\left|\int_0^1a(t)\int_0^tb(s)x(s)dsdt\right|\leq\int_0^1\left|a(t)\int_0^tb(s)x(s)ds\right|dt~~~\mbox{(triangle inequality for integrals)}$$ $$\leq\|a\|\left\|\int_0^tb(s)x(s)ds\right\|~~~\mbox{(Hölder inequality)}$$ $$\leq\|a\|\|b\|\|x\|~~~\mbox{(from previous result)}.$$ Therefore, by definition, $f$ is bounded on $L_2[0,1]$.

I'm stuck on "find an element $y$...".

• Did you try Cauchy-Schwarz? – user71352 Jul 6 '17 at 22:16
• Show us some work. Surely you have some ideas. – zhw. Jul 6 '17 at 22:34
• To find such a $y$ use that $f(x)=(x|y)$ would imply \begin{equation*} \int_{0}^{1}x(s)y(s)ds=\int_{0}^{1}a(t)\int_{0}^{t}b(s)x(s)ds\,dt \end{equation*} then use Fubini's Theorem – user71352 Jul 6 '17 at 23:08
• Riesz Representation Theorem shows that every bounded fanctional functional $f(x)$ on a Hilbert space can be represented as $f(x)=\langle x,y\rangle$. Follow the theorem to find the $y$ you need. – C.Ding Jul 6 '17 at 23:14

The idea is to apply Fubini theorem. The region you are integrating is the triangle $T = \{(s,t)\in \mathbb R^2 | 0 \leq s\leq t \leq 1\}$. By Fubini theorem we have:
$$f(x) = \int_T a(t)b(s)x(s)dsdt = \int_0^1 x(s)b(s)\left(\int_s^1 a(t) dt\right)ds.$$
Therefore your $y$ is given by $$y(s) = b(s)\int_s^1 a(t) dt.$$