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I have to find the operator norm of $A$ as an operator on the Hilbert space $L^2([0,\pi])$, where $A$ is defined as

$$A(f)(x) = \int_0^\pi \sin(x-y) f(y)dy,\text{ where } 0 \le x \le \pi.$$

So the idea I have from lecture notes is to use the following

Theorem:

$E$ Hilbert space and $T$ is self-adjoint operator on $E$. Then we have

$$\| T \| = \sup_{\|x\| = 1}| \langle Tx, x \rangle |.$$

So I think that we have to show is $A = A^*$.

Let $f, g \in L^2([0, \pi])$. then we have the following $$ \begin{align} \langle A(f), g \rangle_{L^2} &= \int_0^\pi \big(A(f)(x)\big)\overline{g(x)}dx \\ &= \int_0^\pi \big( \int_0^\pi \sin(x - y)f(y) dy \big)\overline{g(x)} dx\\ &= \int_0^\pi \int_0^\pi \sin(x - y) f(y) \overline{g(x)} dy dx\\ &= \int_0^\pi f(y) \int_0^\pi \sin(x - y) \overline{g(x)} dx dy\\ &= \int_0^\pi f(y) \int_0^\pi \overline{ \overline{ \sin(x - y) } g(x) dx } dy \\ &= \langle f, A^*(g) \rangle_{L^2} \end{align} $$

Here we have that $k(x, y) = \sin(x - y)$ defined on $(x, y) \in [0, \pi] \times [0, \pi]$ and $ A = A^* $ if $ \overline{k(y, x)} = k(x, y).$ I'm not sure how we are suppose to show this. However, assuming this holds we still have to calculate $$ \|A\| = \sup_{\|f\| = 1} |\langle A(f), f \rangle_{L^2}| $$

We have, $$ \begin{align} \| A \| &= \sup_{\|f\| = 1} | \langle A(f), f \rangle_{L^2} | \\ &= \sup_{\|f\| = 1}|\int_0^\pi A(f)(x) \overline{f(x)}dx|\\ &= \sup_{\|f\| = 1}|\int_0^\pi \big( \int_0^\pi sin(x - y) f(y) dy \big) \overline{f(x)} dx | \\ &= \sup_{\|f\| = 1} | \int_0^\pi \int_0^\pi \sin(x - y)f(y)\overline{f(x)} dy dx | \end{align} $$ but I'm not sure how to tackle this, I haven't got many examples nor experience to solve these types of problems. Maybe there is an another way?

Update: By the hint in the comments I went reading in the course literature. So here is a new idea to solve this:

We find the eigenvalues of $A$: i.e. for a complex number $\lambda$, called the eigenvalue of $A$ if there exist a vector $\mathbf{0} \not = v \in E$ such that $$ Av = \lambda v $$

Now consider the integral operator as in above, we have the following: $$ (Av)(x) = \int_0^\pi \sin(x - y)v(y) dy = \lambda v(x) $$ now since $\sin(x-y) = \sin x \cos y - \cos x \sin y$ we have that $$ (Av)(x) = \sin x \int_0^\pi \cos y \cdot v(y) dy - \cos x \int_0^\pi \sin y \cdot v(y) dy = \lambda v(t) $$ This means that, for $\lambda \not = 0$, $v$ is a linear combination of cosine and sine functions. That is $$ v(x) = a \sin x - b \cos x $$ where $a, b \in \mathbb{C}$. We have that $\dim \mathcal{R}(A) = 2$ where span $\mathcal{R}(A) = \{ \sin x, -\cos x \}$

Now check this out $$ A(-\cos x) = \sin x \underbrace{\int_0^\pi -(\cos^2 y) dy}_{ = - \frac{\pi}{2}} - \cos x \underbrace{\int_0^\pi -\sin y \cos y dy}_{= 0} = - \frac{\pi}{2}\sin x $$ and the same calculation give us that $A(\sin x) = - \frac{\pi}{2} \cos x$ So $A = -\frac{\pi}{2} I$ on the subspace spanned by $\mathcal(R) = \{\sin x, -\cos x\}$ Now, still assuming that $A$ is self-adjoint there is this theorem that states

Theorem: $A$ bounded and self-adjoint on a Hilbert space $\Rightarrow \| A \| = \sup \{ |\lambda| : \lambda \in \sigma(A) \}$

Hence, $\| A \| = \frac{\pi}{2}.$

Is this solution somewhat correct reasoning?

Supersalt
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1 Answers1

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Hint

$A$ has rank 2 here, $V := \text{Im}(A)$ being spanned by the two orthogonal functions $u(x) = \cos x$ and $v(x) = \sin x$. Observe that $A$ vanishes on $V^\bot$ and conclude that the maximum of $\|Af\|$ when $\|f\| = 1$ is reached when $f\in V$. You are left with the problem of computing the norm of an operator in a 2-dimensional space. Write $f = \lambda u + \mu v$ with $\lambda^2 + \mu^2 = C$, compute and maximize $\|A f\|$

Gribouillis
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  • That's certainly correct. Let's hope the OP will find it helpful. –  Oct 07 '17 at 14:19
  • Thanks! This did indeed make my problem easier. I have updated the original post on what I came up with reading in the course literature. However, I still haven't shown that $k(x, y) = \overline{k(y, x)}$ because then the operator $A$ will be self-adjoint though. Is this hard to show? – Supersalt Oct 07 '17 at 17:07
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    @Supersalt $A$ isn't self-adjoint, the operator $B$, defined the same way, but with $\cos(x-y)$ as kernel, is. That's because $P=\frac2{\pi}B$ is the orthogonal projection on $V$, so $P$ is self-adjoint. Interestingly, $|Bf|=|Af|$, so it's clear $|A|=|B|=\frac{\pi}2|P|=\frac{\pi}2$. Still more interesting: $A=-DB$, where $D$ is differentiation. That's usually not bounded, but on $V$, it is, and has norm $1$. –  Oct 07 '17 at 18:27
  • @ProfessorVector $A$ is skew-adjoint. Its restriction to the plane spanned by the orthonormal pair $c(x)=\sqrt{\frac{2}{\pi}}\cos(x) , s(x)=\sqrt{\frac{2}{\pi}}\sin(x)$ is a rotation followed by a dilation of ratio $\frac{\pi}{2}$. One has $A f =\frac{\pi}{2}\langle f, c\rangle s -\frac{\pi}{2}\langle f, s\rangle c $ for all $f$. – Gribouillis Oct 07 '17 at 19:46
  • Yeah, I know, I just dared call what you name "rotation" differentiation. It is, in this case. –  Oct 07 '17 at 19:56