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Is it true that $$|| a R+b P||\leq\max \{|a|,|b|\},$$where $a$ and $b$ are complex numbers and $P,R$ are (orthogonal) projection operators on finite-dimensional closed subspaces of an infinite-dimensional hilbert space ?

In finite dimensions this would be true, since a projection there has always norm one.

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2 Answers 2

up vote 6 down vote accepted

It is true when $R,P$ are orthogonal to each other (there is an ambiguity in terminology, as "orthogonal" could mean that $PR=0$, or that $P=P^*=P^2$, $R=R^*=R^2$.

If $PR=0$ is not assumed, then the answer is no: take $P=R=I$, $a=b=1$, then $\|aR+bP\|=2$.

Assuming $PR=0$, then $\|aR+bP\|=\max\{|a|,|b|\}$. Indeed, for any $\xi$ in the range of $R$ with $\|\xi\|=1$, we have $$ \|(aR+bP)\xi\|=\|aR\xi\|=\|a\xi\|=|a|; $$ similarly, $\|(aR+bP)\eta\|=|b|$ if $\eta$ is in the range of $P$ and $\|\eta\|=1$. So $\|aR+bP\|\geq\max\{|a|,|b|\}$.

For an arbitrary vector $\nu$ with $\|\nu\|=1$, we can write $\nu=\alpha\nu_1+\beta\nu_2+\gamma\nu_3$, for three unit vectors with $\nu_1$ in the range of $R$, $\nu_2$ in the range of $P$, and $\nu_3$ orthogonal to both the ranges of $R$ and $P$, and $\alpha,\beta,\gamma\geq0$, $\alpha^2+\beta^2+\gamma^2=1$ (see edit below for an explanation). Then $$ \|(aR+bP)\nu\|^2=\|a\alpha\nu_1+b\beta\nu_2\|^2=|a|^2\,\alpha^2 + |b|^2\,\beta^2\leq\max\{|a|^2,|b|^2\}, $$ so $\|(aR+bP)\nu\|=\sqrt{\alpha^2|a|^2+\beta^2|b|^2}\leq\max\{|a|,|b|\}$.

In conclusion, $\|aR+bP\|=\max\{|a|,|b|\}$.

Edit: below is a proof of the claim that, given three pairwise orthogonal subspaces $X$, $Y$, $Z$ of a Hilbert space $H$ that span the whole space, any vector $\nu\in H$ can be written as $$\nu=\alpha\nu_1+\beta\nu_2+\gamma\nu_3,\ \ \ \nu_1\in X,\ \nu_2\in Y, \ \nu_3\in Z,$$ with $\alpha,\beta,\gamma\geq0$ and $\alpha^2+\beta^2+\gamma^2=1$.

Let $\{x_j\}$, $\{y_j\}$, $\{z_j\}$ be orthonormal bases for $X$, $Y$, $Z$. Together, they form a basis for the whole $H$. So there exist coefficients such that $$ \nu=\sum_j a_jx_j + \sum_jb_jy_j+\sum_jc_jz_j. $$ As $\|\nu\|=1$, $\sum_j|a_j|^2+\sum_j|b_j|^2+\sum_j|c_j|^2=1$. Let $$ \alpha=(\sum_j |a_j|^2)^{1/2},\ \beta=(\sum_j |b_j|^2)^{1/2},\ \gamma=(\sum_j |c_j|^2)^{1/2},\ $$ and $$ \nu_1=\sum_j\frac{a_j}\alpha\,x_j,\ \nu_2=\sum_j\frac{b_j}\beta\,y_j,\ \nu_3=\sum_j\frac{c_j}\gamma\,z_j. $$ Then $\nu_1,\nu_2\nu_3$ are unit vectors, $\alpha^2+\beta^2+\gamma^2=1$, and $$ \nu=\alpha\nu_1+\beta\nu_2+\gamma\nu_3. $$

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1  
Careful, $\sqrt{|a|^2+|b|^2} > \max\{|a|,|b|\}$, unless $ab=0$. –  Erick Wong Jul 29 '12 at 18:03
    
Thanks, I forgot to use the convex coefficients. It's corrected now. –  Martin Argerami Jul 29 '12 at 23:23
    
@MartinArgerami Before I accept your answer could you please explain, why "For an arbitrary vector $\nu$ with [.....] $\alpha+\beta+\gamma=1$" holds ? (Especially, why is $\alpha+\beta+\gamma=1$?) –  user36772 Jul 30 '12 at 10:00
    
That was a small mistake. Actually, their squares add to $1$. I've corrected the mistake and added the explanation. –  Martin Argerami Jul 30 '12 at 15:37
    
I think the detour through bases is unnecessary; also one of $\alpha$, $\beta$, $\gamma$ might be zero. See my answer to this question. –  joriki Aug 24 '12 at 4:32

Let $\|x\|\le 1$. Then $$ \begin{eqnarray} \|aRx + bPx\|^2 &=& |a|^2\|Px\|^2 + |b|^2\|Rx\|^2\\ &\le& (|a|^2\wedge |b|^2)(\|Px\|^2 +\| Qx\|^2)\\ & \le& (|a|^2\wedge |b|^2)\|x\|^2. \end{eqnarray} $$

You can now conclude your result. The important thing is that $P$ and $Q$ are orthogonal so $\|Px + Qx\|^2 \le \|x\|^2.$

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