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Let $(X,\langle\cdot ,\cdot\rangle)$ an inner product space and $A\in\mathcal L(X)$. I have to show that $$\|A\|=\sup\{|\langle Ax,x\rangle|\mid x\in X, \|x\|\leq 1\}.$$ The fact that $\|A\|\geq \sup\{|\langle Ax,x\rangle|\mid x\in X, \|x\|\leq 1\}$ is a consequence of Cauchy-Schwarz. For the other inequality, we set $m=\inf\{\langle Ax,x\rangle\mid x\in X, \|x\|=1\}$ and $M=\sup\{\langle Ax,x\rangle\mid x\in X, \|x\|=1\}$. So we want to prove that $\|A\|\leq \max\{-m,M\}$.

Q1) Why $m\leq 0$ ?

Moreover, in the proof, we have $$\max\{-m,M\}=\sup\{|\langle Ax,x\rangle|\mid x\in X,\|x\|=1\}\underset{(*)}{=}\sup\{|\langle Ax,x\rangle|\mid x\in X,\|x\|\leq 1\}.$$

Q2) Why $(*)$ is an equality ? Shouldn't it be a $\leq$ ? I asked this question to my teacher, and he told me that it was correct but with no explanation.

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    $\begingroup$ $m \le 0$ because $\langle A0, 0 \rangle = 0$. $\endgroup$
    – Crostul
    Jun 19 '15 at 14:39
  • $\begingroup$ Is $\|\cdot\|$ the operator norm? Is $A$ Self-adjoint? $\endgroup$ Jun 19 '15 at 14:39
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    $\begingroup$ It seems like you are comparing the operator norm to the numerical radius. The two are not equal for arbitrary operators. $\endgroup$ Jun 19 '15 at 14:41
  • $\begingroup$ $\| A\|$ is the norm of the operator $A$, and $\|x\|$ the norm of the element $x$. $\endgroup$
    – idm
    Jun 19 '15 at 14:41
  • $\begingroup$ For unbounded normal operators the norm agrees with the numerical radius. That is a consequence of: $N^*N=NN^*:\quad\langle\sigma(N)\rangle=\overline{\mathcal{W}(N)}\quad$ Apart from that case there's not too much to hope for. However inclusion remains for bounded operators: $A\in\mathcal{B}(\mathcal{H}):\quad\langle\sigma(A)\rangle\subseteq\overline{ \mathcal{W}(A)}$ $\endgroup$ Jun 19 '15 at 17:04
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Q1) This is false, take $A=Id$, then $\langle x, x\rangle =\|x\|^2$ for all $x$. So $m=\inf\{\langle Ax,x\rangle\mid x\in X, \|x\|=1\}$ can be positive.

Q2) If $\|x\|\leq 1$, then we can write $\langle Ax, x\rangle =\|x\| ^2\langle \frac{Ax}{\|x\|}, \frac{x}{\|x \|}\rangle \leq \sup\{|<Ay,y>|\mid y\in X, \|y\|=1 \}$ since $\|x\|^2\leq 1$. The converse inequality is obvious.

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  • $\begingroup$ Thanks. So why do we want to prove that $\|A\|\leq \max\{-m,M\}$ and not that $\|A\|\leq\max\{M,m\}$ ? $\endgroup$
    – idm
    Jun 19 '15 at 14:52
  • $\begingroup$ Because if $m\leq y \leq M$ then $|y|\leq \max (|m|, M)$. If $m>0$ it is clear that $\max (-m,M)=\max (|m| ,M)$ by definition of $m$ and $M$. If $m\leq 0$ we also have $\max (-m,M)=\max (|m| ,M)$. $\endgroup$
    – Patissot
    Jun 19 '15 at 15:00

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