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Given two real numbers $\alpha$ and $\beta$, consider the linear operator $T:\mathbb{C}\rightarrow \mathbb{R}$ defined by $T(x+iy)=\alpha x +\beta y$.

I am trying to figure out the norm of this operator:

$$\| T\|=\sup\limits_{z \in \mathbb{C}\neq 0}\left\{\frac{\|Tz \|}{\| z\|} \right\}=\sup\limits_{(x,y)\neq (0,0)}\left\{ \left| \frac{\alpha x +\beta y }{ x+iy}\right|\right\}=\sup\limits_{(x,y)\neq (0,0)}\left\{ \frac{\left|\alpha x +\beta y\right| }{ \sqrt{x^2+y^2}}\right\} $$

I thought maybe using polar coordinates to simplify the ratio as follows:

$$ \frac{\left|\alpha x +\beta y\right| }{ \sqrt{x^2+y^2}}=\frac{\left|\alpha r \cos(\theta) +\beta r \sin(\theta)\right| }{ r}= \left|\alpha \cos(\theta) +\beta \sin(\theta)\right| $$

By using derivatives, this function reaches its maximum when $\tan{\theta}=\frac{\beta}{\alpha}$, i.e. : $$ \| T\| = \left|\alpha \cos(\arctan{\frac{\beta}{\alpha}}) +\beta \sin(\arctan{\frac{\beta}{\alpha}})\right|=\left| \alpha\frac{1}{\sqrt{1+\left(\frac{\beta}{\alpha}\right)^2}}+\beta\frac{ \frac{\beta}{\alpha}}{\sqrt{1+\left(\frac{\beta}{\alpha}\right)^2}} \right|=\sqrt{\alpha^2+\beta^2} $$

Any other approach is welcome.

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  • $\begingroup$ I realize the problem is easier to solve by using another different of $\| T\|$ (as Shalop did), namely $ \| T \|=\sup\limits_{z \in \mathbb{C}}\left\{ \|Tz \|\;|\;\| z\|=1 \right\}$: the solution is at the intersection between the circle of radius one centered at (0,0), and the line $\alpha x +\beta y$…very nice resolution method with Cauchy Schwartz proposed by Shalop. $\endgroup$ – Kuifje Nov 6 '15 at 23:34
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If $x^2+y^2=1$, then by Cauchy Schwarz, $|\alpha x + \beta y| \leq (\alpha^2+\beta^2)^{1/2}$.

On the other hand, $T\big(\frac{\alpha}{(\alpha^2+\beta^2)^{1/2}}+i\frac{\beta}{(\alpha^2+\beta^2)^{1/2}}\big)= (\alpha^2+\beta^2)^{1/2}$.

So, $\|T\| = (\alpha^2+\beta^2)^{1/2}$.

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    $\begingroup$ We could add that the Cauchy Schwartz is an equality when $(\alpha,\beta)$ and $(x,y)$ are collinear, i.e. when $(x,y)=k(\alpha,\beta)$, and the constraint $\| z\|=1 $ imposes $k=1/\sqrt{\alpha^2+\beta^2}$. $\endgroup$ – Kuifje Nov 6 '15 at 23:52

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