Compute the adjoint and norm of a integral operator in $L_2[0,1]$

Question

Let $A:L_2[0,1]\to L_2[0,1]$ be defined by$$Ax(t)=\int \limits _0^1t\left (s-\frac{1}{2}\right )x(s)\,ds\quad \forall t\in [0,1].$$Compute the adjoint and the norm of $A$

This is my avance:

First I think to $\|A\|=\dfrac{1}{2}$ I only prove that:

Let $x\in L_2[0,1]$ then$$\|Ax(t)\|_2=\left (\int \limits _0^1t\left (s-\frac{1}{2}\right )x(s)\,ds\right )^{1/2}\leq \frac{1}{2}\|x\|_2.$$Is this correct?
For the other inequality I try to find $x$ with norm $1$ such that $\|Ax\|=\dfrac{1}{2}$.
Any hint or help for this step or for computing the adjoint will be greatly appreciated.

Answer 1

The operator is of the form $$ Ax=\langle x, g\rangle \,f$$ where $f(t)=t$ and $g(t)=t-{1\over 2}.$ Thus the range is one-dimensional. Its norm is equal $$\|A\|=\|f\|_2\|g\|_2={1\over \sqrt{3}}{1\over \sqrt{12}}={1\over 6}$$ The adjoint operator can be calculated as follows $$\langle x,A^*y\rangle =\langle Ax,y\rangle=\langle x,g\rangle \langle f,y\rangle=\langle x,\langle y,f\rangle g\rangle $$ Hence $$A^*y =\langle y,f\rangle g$$ i.e. the functions $f$ and $g$ swapped. The operator $A$ is not self-adjoint if $f$ and $g$ are linearly independent, like in the OP question. Explicitly we get $$(A^*x)(t)=\left (t-{1\over 2}\right )\int\limits_0^1 s\,x(s)\,ds$$ Remark For operator of the form $$(Ax)(t)=\int\limits_0^1k(s,t)x(s)\,ds$$ the adjoint operator is given by

$$(A^*x)(t)=\int\limits_0^1\bar{k}(t,s)x(s)\,ds$$ Thus $A^*=A$ if and only if $k(s,t)=\bar{k}(t,s).$ In the OP question we have $k(s,t)=t(s-{1\over 2}), $ hence $k(s,t)\neq k(t,s).$

Answer 2

Not quite. I'll tell you what you did wrong first and try to point you in the right direction. My solution will be presented later and you can use that as a reference.

So, your calculation of $\left|\left|Ax\right|\right|_2$ is incorrect. You haven't used the correct expression. We have the map: $$t \mapsto Ax(t) := t \int_{0}^{1} (s-\frac{1}{2})x(s) \ ds$$ The correct formula for the $L^2$-norm of $Ax$ is: $$\left|\left|Ax\right|\right|_2 = \left(\int_{0}^{1} t^2 \left|\int_{0}^{1} \left(s-\frac{1}{2}\right)x(s) \ ds \right|^2 \ dt \right)^{\frac{1}{2}}$$

Start with this and begin simplifying. You might possibly have to use the Cauchy-Schwarz Inequality along the way and justify why that gives you the best possible estimate. As for the adjoint, you need to start by playing around with the following equation: $$\langle Tf,g \rangle = \langle f,T^{\star} g \rangle$$

The adjoint is defined by this and if you play around with it, you'll get an expression for what the adjoint should be.

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Solution:

Observe that: $$\left|\left|Ax\right|\right|_2^2 = \frac{1}{3} \left|\int_{0}^{1} \left(s-\frac{1}{2} \right) x(s) \ ds \right|^2$$ By the Cauchy-Schwarz Inequality, we have that: $$\left|\int_{0}^{1} \left(s-\frac{1}{2}\right) x(s) \ ds \right| \leq \left(\int_{0}^{1} \left(s-\frac{1}{2} \right)^2 \ ds \right)^{\frac{1}{2}} \left|\left|x\right|\right|_2 = \frac{1}{2\sqrt{3}} \left|\left|x\right|\right|_2$$ I've left the calculation of the integral to you, that should be pretty easy. Therefore, it follows that: $$\left|\left|Ax\right|\right|_2^2 \leq \frac{1}{3} \cdot \frac{1}{12} \cdot \left|\left|x\right|\right|_2^2$$ Hence, it follows that: $$\left|\left|Ax\right|\right|_2 \leq \frac{1}{6} \left|\left|x\right|\right|_2$$ as was desired. Now, I claim that $\left|\left|A\right|\right| = \frac{1}{6}$. To check that this is the case, let $x(s) = s-\frac{1}{2}$. Then: $$\left|\left|x\right|\right|_2 = \left(\int_{0}^{1} \left(s-\frac{1}{2} \right)^2 \ ds \right)^{\frac{1}{2}} = \frac{1}{2\sqrt{3}}$$ Now, define $x_0(s) = 2\sqrt{3} (s-\frac{1}{2})$. This has $L^2$-norm equal to $1$ based on the calculation we just did. Observe that: $$Ax_0(t) = t \int_{0}^{1} 2\sqrt{3}(s-\frac{1}{2})^2 \ ds = \frac{t}{2\sqrt{3}}$$ Therefore, we have that: $$\left|\left|Ax_0 \right|\right|_2 = \left(\int_{0}^{1} \frac{t^2}{12} \ dt \right)^{\frac{1}{2}} = \frac{1}{6}$$ In other words, we knew that $\left|\left|A\right|\right| \leq \frac{1}{6}$ and it turns out that this map actually "achieves" the bound of $\frac{1}{6}$. It follows that $\left|\left|A\right|\right| = \frac{1}{6}$.

Next, let's calculate the adjoint. For this, let's simplify our notation. Let $k(s,t) = t(s-\frac{1}{2})$. Then: $$Ax(t) = \int_{0}^{1} k(s,t) x(s) \ ds$$ Therefore, we have that: $$\langle Ax, y \rangle = \int_{0}^{1} \overline{Ax(t)} y(t) \ dt = \int_{0}^{1} \int_{0}^{1} k(s,t) \overline{x(s)} y(t) \ ds \ dt$$ $$\langle Ax,y \rangle = \int_{0}^{1} \overline{x(s)} \int_{0}^{1} k(s,t) y(t) \ dt \ ds = \langle x, Ay \rangle $$

This actually proves that the operator is symmetric. Now, it follows that the operator is self-adjoint and we have that: $$A^{\star} = A$$

as was desired.

Edit:

So, the calculation above for the adjoint is incorrect. The idea behind it is correct but what I derived towards the end isn't. Have a look at Ryszard's answer if you want the correct derivation, I might correct mine later. But everything else should be quite fine.