Norm of an integral operator in $L^2$

Question

Let T an operator in $L^2((-1,1))$ such that $Tf(x) = \int_{-1}^{1} (xy + y^2)f(y)dy$.

Calculate the norm of $T$.

I tried calculating $\left\lVert Tf(x) \right\rVert$ first, but I got stucked with the square of the inside integral. Any help?

Answer 1

As known, $\|T\|^2=\|TT^\ast\|$, where $T^\ast f=\displaystyle\int\limits_{-1}^1(xy+x^2)f(y)dy$.

Since the operator $TT^\ast f=\displaystyle\int\limits_{-1}^1\left(\dfrac{2}{5}+\dfrac{2}{3}xy\right)f(y)dy$ is self-adjoint and compact, its norm is equal to the maximum modulo eigenvalue. Consider the equation $\dfrac{2}{5}\displaystyle\int\limits_{-1}^1f(y)dy+\dfrac{2}{3}x\displaystyle\int\limits_{-1}^1yf(y)dy=\lambda f(x)$. We can assume that $\lambda\ne0$, therefore $f(x)=c_1+c_2x$, where $c_1^2+c_2^2\ne 0$. Substituting $f(x)$ into the equation, we obtain the system $\lambda c_1-\dfrac{4}{5}c_1=0$, $\lambda c_2-\dfrac{4}{9}c_2=0$, where do we find two eigenvalues $\lambda_1=\dfrac{4}{5}$, $\lambda_1=\dfrac{4}{9}$. Finally, $\|A\|=\dfrac{2}{\sqrt5}$.

Answer 2

Observe \begin{align} Tf(x) = x\int^1_{-1} dy\ yf(y)+\int^1_{-1}dy\ y^2f(y)=: \mu x +v \end{align} then we see that \begin{align} \|Tf\|_{2}^2 = \int^1_{-1} dx\ |\mu x+v|^2 = \int^1_{-1}dx\ \mu^2x^2+\mu v x + v^2= \frac{2\mu^2}{3}+2v^2. \end{align} Next, observe that for any $f \in L^2(-1, 1)$, we can write \begin{align} f(x) = \sum^\infty_{n=0} c_n P_n(x) = \frac{c_0}{\sqrt{2}}+ c_1\sqrt{\frac{3}{2}}x+c_2\sqrt{\frac{5}{8}}(3x^2-1)+\cdots \end{align} where $P_n$ is the normalized $n$th degree Legendre polynomial. Hence it follows \begin{align} \|f\|_2^2= \sum^\infty_{n=0}|c_n|^2 \end{align} with \begin{align} \mu =\int^1_{-1} dy\ yf(y) =\sqrt{\frac{2}{3}}c_1 \end{align} and \begin{align} v=\int^1_{-1} dy\ y^2f(y) = \frac{\sqrt{2}}{3}c_0+\frac{2}{3}\sqrt{\frac{2}{5}}c_2. \end{align}

Finally, we see that \begin{align} \|Tf\|_2^2=&\ \frac{2}{3}\left(\sqrt{\frac{2}{3}}c_1 \right)^2+2\left(\frac{\sqrt{2}}{3}c_0+\frac{2}{3}\sqrt{\frac{2}{5}}c_2 \right)^2\\ \le&\ \frac{4}{9}c_1^2+2\left(\frac{2}{9}+\frac{8}{45}\right)(c_0^2+c_2^2)\\ \leq&\ 2\left(\frac{2}{9}+\frac{8}{45}\right)(c_0^2+c_1^2+c_2^2)= \frac{4}{5}(c_0^2+c_1^2+c_2^2) \end{align} which means \begin{align} \|T\|_\text{op} \leq \sqrt{\frac{4}{5}}\approx 0.894. \end{align}

Consider $f(x) = P_0(x)+cP_2(x)$, then we see that \begin{align} \|Tf\|_2^2=2\left(\frac{\sqrt{2}}{3}+\frac{2}{3}\sqrt{\frac{2}{5}}c \right)^2=\frac{4}{5}(1+c^2) = \frac{4}{5}\|f\|_2^2 \end{align} provided $c=\frac{2}{\sqrt{5}}$. Hence $\|T\|_\text{op} = \sqrt{4/5}$.