# Prove that the following operator is bounded, self-adjoint and compact.

Let $$(Tf)(x) = \int_{0}^{1} \sinh(x-t)f(t)dt$$ be an operator on $L^2\left([0,1]\right)$.

Show that it is bounded, self-adjoint and compact. Find its norm.

As I understand, if I prove that it is self-adjoint and compact, then I can try to find its eigenvalues, because $\| T \| = \sup |\lambda|$.

I tried to show, that $T$ is bounded by definition: $$\|Tf\|^2 = \int_{0}^{1} \left| \int_{0}^{1} \sinh(x-t)f(t) dt \right|^2 dx \ .$$ However, I have no idea how to get $C \|f\|$ on the right-hand sight.

Compactness imply boundedness, but I don't see how can I prove it too. And the same with self-adjointness.

• Use Fubini's theorem, and try to get the constant (which will be the integral of the hyperbolic sine) out. This is called a convolution operator, and it has many nice properties. – астон вілла олоф мэллбэрг Jan 24 '17 at 0:08
• @астонвіллаолофмэллбэрг if I can change the order of integration, then $\| Tf \| \leq \| \sinh \| \| f \|$ (if I did it right). But I still have no ideas about compactness and self-adjointness. – nightked Jan 24 '17 at 0:16
• Otherwise, you can look at the Fourier series of $f$ and $\sinh$ on $[0,2]$, the (circular) convolution transforming nicely under Fourier series. For the adjoint, start from $\langle Tf, g \rangle =\int\int \ldots$ and find $S$ such that it is $=\langle f, Sg \rangle$ – reuns Jan 24 '17 at 0:18
• For self-adjointness, try to see what are the barriers you face when you write $\langle Tf, g \rangle = \langle f , Tg \rangle$. Just a look at both forms will tell you what are the missing pieces. For compactness, take a bounded sequence $f$, and try to use dominated convergence theorem. – астон вілла олоф мэллбэрг Jan 24 '17 at 0:20
• I haven't checked yet, but I don't think that $T$ is self adjoint. $\sinh (x-t) = - \sinh(t-x)$. – copper.hat Jan 24 '17 at 0:29

Note that $$\sinh(x-t)=\sinh(x)\cosh(t)-\cosh(x)\sinh(t)$$ Therefore your operator $T$ is rank 2 operator, which makes it bounded and compact: $$Tf = (f,c)s-(f,s)c.$$ And $T$ is anti-symmetric because $$(Tf,g) = ((f,c)s-(f,s)c,g)=(f,c)(s,g)-(f,s)(c,g) \\ (f,Tg) = (f,(g,c)s-(g,s)c)=(c,g)(f,s)-(s,g)(f,c) \\ (Tf,g) = -(f,Tg).$$ The norm is reduce to that of a two-dimensional operator.
Cauchy Schwartz gives $|Tf(x)|^2 \le \int_I \sinh^2(x-t) dt \|f\|^2$ and so $\|Tf\| \le \sqrt{\int_I \int_I \sinh^2(x-t) dt dx} \|f\|$ (this is a Hilbert Schmidt operator).
Note that \begin{eqnarray} \langle g , T f \rangle &=& \int_I g(x) (Tf)(x) dx \\ &=& \int_I \int _I g(x) \sinh (x-t) f(t) dt dx \\ &=& \int_I \int _I g(t) \sinh (t-x) f(x) dt dx \\ &=& -\int_I \int _I g(t) \sinh (x-t) f(x) dt dx \\ &=& -\langle Tg , f \rangle \end{eqnarray} and so $T^* = -T$.
To show that it is compact, one standard approach is to show that $T$ is the limit of finite rank operators. To do this, one can get the Fourier expansion of $K(x,y) = \sinh (x-y)$ in terms of the basis functions $e_{mn}(s,t)= e_m(s) e_n(t)$, where $e_k(\omega) = e^{i 2 k\pi \omega }$ and approximate $K$ by a truncated Fourier series. The resulting operators have finite rank and converge to $T$.