# Norm of two operators

(1) $U(x)=a \langle x,b \rangle +b \langle x,a \rangle$, $a,b\in H\setminus \{0\}$. U is an operator from H to H and a,b are orthogonal elements. I want to calculate $||U||$

For this one I tried the following:

$||U(x)||^2=||a \langle x,b \rangle +b \langle x,a \rangle ||^2=| \langle x,b \rangle |^2||a||^2+| \langle x,a \rangle |^2||b||^2\le2||a^2|| |b||^2 ||x||^2$, by using Cauchy Schwarz

=> $||U||\le \sqrt{2}||a|| ||b||$ ? When do I have equality here and how can I derive $||U||=$ from it?

(2)$U:L^2([0,\pi])->L^2([0,\pi])$ defined by $U f(x)=\sin(x)\int_{0}^{\pi}f(t)\cos(t)dt+\cos(x)\int_{0}^{\pi}f(t)\sin(t)dt$

Here I have no idea how to derive $||U||$

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You can think of $a/\|a\|,b/\|b\|$ as the first two elements in an orthonormal basis. Then $$U=\|a\|\,\|b\|\,\begin{bmatrix} 0&1\\1&0\end{bmatrix}$$ As the $2\times 2$ matrix is a unitary (so norm equal to $1$), $\|U\|=\|a\|\,\|b\|$.
If you want to make this into a formal proof, using Cauchy-Schwarz in the inner products in your formula is too crude. You can do the following: $$\|U(x)\|^2=\|a \langle x,b \rangle \|^2+b \langle x,a \rangle \|^2=| \langle x,b \rangle |^2\|a\|^2+| \langle x,a \rangle |^2\|b\|^2=| \langle x,\frac{b}{\|b\|} \rangle |^2\|b\|^2\,\|a\|^2+| \langle x,\frac{a}{\|a\|} \rangle |^2\|a\|^2\,\|b\|^2 =\|a\|^2\,\|b\|^2\,(| \langle x,\frac{b}{\|b\|} \rangle |^2+| \langle x,\frac{a}{\|a\|} \rangle |^2)\leq\|a\|^2\|b\|^2\,\|x\|^2.$$ So $\|U\|\leq\|a\|\,\|b\|$. The equality is achieved at both $a$, and $b$, so $\|U\|=\|a\|\,\|b\|$.
Also note that if $v \in \text{sp}\{a,b\}^\bot$, then $U(x+v) = U(x)$. So, you need only consider the two dimensional subspace $\text{sp}\{a,b\}$ when computing the norm $\|U\|$. – copper.hat Nov 21 '12 at 17:29
The second one is a particular case of the first one: $\sin$ and $\cos$ are orthogonal in $L^2[0,\pi]$. – Martin Argerami Nov 21 '12 at 17:40
Exactly.${\ \ }$ – Martin Argerami Nov 21 '12 at 17:49