The following is a question I have been working on for some time with help from my teacher. Unfortunately we have a solution but are not 100% confident with it. Some guidance on if part(s) of our solution are incorrect would be greatly appreciated. Any help is most welcome. Thankyou!
Show that B is a bounded linear operator on $L^{2}_{(R_+)}$. Calculate the adjoint operator.
Set $R_+=\{x \in R:x \ge 0\}$ For $f \in L^{2}(R_+)$. Define: $$Bf(x)=\dfrac{1}{x}\int_{0}^{x}f(t) \ dt, x>0.$$
Showing boundedness. Assume $Bf(x)$ is linear (Shown later).
We wish to show that this is continuous in $0$. By definition, we have that $x$ is dense in $L^{2}_{(R_+)}$. We also know that $f_n$ is uniformly bounded (Unsure on this part?)
Let $f_{n} \in C^{\infty}_{c}(R_+)$
Define $F_{n}(x):=\dfrac{1}{x^{2}} \lvert\int_{0}^{x}f_{n}(t) \ dt\rvert^{2}$ with $F_{n}(x) \in L^2(R_+)$
Then we know $F_{n}(0)\equiv 0 \ \ \ \forall n \ \ As \ f_{n} \to 0$
Claim: $\lvert\lvert(F_{n}(x))\rvert\rvert_{L^2_{(R_+)}}\to 0$
\begin{align} \lvert\lvert(F_{n}(x))\rvert\rvert_{L^2_{(R_+)}}&=[\int_{0}^{\infty}\lvert\dfrac{1}{x^{2}}\lvert\int_{0}^{x}f_n(t) \ dt\rvert^{2}\rvert^{2} dx]^{1/2} \\ &=\int_{0}^{1}\dfrac{1}{x^{2}}\lvert\int_{0}^{x}f_n(t) \ dt\rvert^{2} dx+\int_{1}^{\infty}\dfrac{1}{x^{2}}\lvert\int_{0}^{x}f_n(t) \ dt\rvert^{2} dx \end{align}
For \begin{align} \int_{0}^{1}\dfrac{1}{x^{2}} \ dx \to 0 \ As \ F_{n}(0) \equiv 0 \end{align} \begin{align} \lvert\int_{0}^{x}f_n(t) \ dt\rvert^{2} \le \int_{0}^{x}\lvert f_n(t)\rvert^{2} \ dt \to 0 \ As \ \ \ f_{n}(t) \to 0 \ \ (Holder's \ Inequality) \end{align}
For $\int_{1}^{\infty}\dfrac{1}{x^{2}}\lvert\int_{0}^{x}f_n(t) \ dt\rvert^{2} \ dx$ we have that:
\begin{align} \int_{1}^{\infty}\dfrac{1}{x^{2}} \ dx \le M \end{align} \begin{align} \lvert\int_{0}^{x}f_n(t) \ dt\rvert^{2} \le \int_{0}^{x}\lvert f_n(t)\rvert^{2} \ dt \to 0 \ As \ \ \ f_{n}(t) \to 0 \ \ (Holder's \ Inequality) \end{align}
So $f_{n} \in L^{2}_{(R_+)}$
Hence, by density, $f_{n} \to f$ as $n \to \infty$, and we have that: \begin{align} \lvert\lvert(Bf(x))\rvert\rvert_{L^2_{(R_+)}}&=(\int_{0}^{\infty}\lvert\dfrac{1}{x} \ \int_{0}^{x}f(t) \ dt\rvert^{2} \ dx)^{1/2} \\ &=(\int_{0}^{1}\lvert\dfrac{1}{x} \ \int_{0}^{x}f(t) \ dt\rvert^{2} \ dx)^{1/2} +(\int_{1}^{\infty}\lvert\dfrac{1}{x}\int_{0}^{x}f(t) \ dt\rvert^{2} \ dx)^{1/2} \\ &\le \int_{0}^{1}\dfrac{1}{x} \ dx \ \int_{0}^{x}\lvert f(t)\rvert \ dt+\int_{1}^{\infty}\dfrac{1}{x} \ dx \ \int_{0}^{x}\lvert f(t)\rvert \ dt \\ &\le M\lvert\lvert f \rvert\rvert_{L^2_{(R_+)}} \end{align}
Therefore $Bf(x)$ is bounded for some constant $M > 0$.
Linearity: Let $\alpha,\beta \in R_+$. Then $$ \begin{align*} B(\alpha f+\beta f)(x) &=(\dfrac{1}{x}+\dfrac{1}{x})\int_{0}^{x}(\alpha f(t)+ \beta f(t))dt \\ &=(\dfrac{1}{x}\int_{0}^{x}\alpha f(t)dt +\dfrac{1}{x}\int_{0}^{x}\beta f(t)dt) \\ &=\alpha Bf(x)+\beta Bf(x) \end{align*} $$
Therefore $B(f(x))$ is linear.
Adjoint operator:
We need to find $B^{*}$. We know by definition $(Bf,u)_{L^{2}_(R_+)}=(f,B^{*}u)_{L^{2}_(R_+)}$.
Consider the characteristic \begin{align} \chi(t,x) &=\{1 \ \ x > t \\ &= \{0 \ \ x \le t & \end{align}
From above we have:
\begin{align} (B(f),u)_{L^{2}_{(R_+)}} &=\int_{0}^{\infty}[\dfrac{1}{x}\int_{0}^{x} \ f(t) \ dt] \ \chi(t,x) \ u(x) dx \\ &=\int_{0}^{\infty}\dfrac{1}{x}\int_{0}^{x} \ f(t) \ \chi(t,x) \ dt \ u(x) dx \\ &=\int_{0}^{x} \ f(t) \ dt \ \int_{0}^{\infty}\dfrac{1}{x} \ u(x) \ \chi(t,x) \ dx \\ &=\int_{0}^{x} \ f(t) \ dt \ \int_{t}^{\infty}\dfrac{1}{x} \ u(x) \ dx \\ &=(f,B^{*}u) \end{align}
Hence $B^{*}=\int_{t}^{\infty}\dfrac{1}{x} \ dx$
The part I am most unsure on is my solution to $B(f(x))$ at the start. I am unsure I have applied the $L^{2}_{(R_+)}$ norm correctly.
Here's the issue with your boundedness proof. Your argument boils down to the following calculation: \begin{align*} \|Bf\|_{L^2}&=(\int_0^\infty\left|\frac{1}{x}\int_0^xf(t)dt\right|^2dx)^{1/2}\\ &\leq(\int_0^\infty\left(\frac{1}{x}\int_0^\infty|f(t)|\mathbf1_{[0,x]}dt\right)^2dx)^{1/2}\\ &\leq(\int_0^\infty\frac{1}{x^2}\|f\|_{L^2}^2xdx)^{1/2}\\ &=(\int_0^\infty\frac{dx}{x})^{1/2}\|f\|_{L^2}. \end{align*} As you can see, this does not prove boundedness because $1/x$ is not integrable at zero or infinity. The inequality that is not tight is Cauchy-Schwarz, since $|f(t)|$ can never be an exact multiple of $\mathbf1_{[0,x]}(t)$. Therefore a stronger argument is needed to prove boundedness.