Notes on some inequalities for linear operators
Notes on some inequalities for linear operators
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关于线性算子的一些不等式的注释
DOI：
10.1007/bf01343117
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发表时间：
1952
影响因子：
1.4
通讯作者：
Tosio Kato
中科院分区：
文献类型：
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作者：
Tosio Kato
引理岛设T是线性算子，T* 和T-1存在.则T* 有唯一的右逆T *“，具有下列性质2）：i）~ T*"=~"*，ii）~ T *”G ~ T，iii）~ T *“~）T* 和T* T *”= I in ST.如果特别地T* 具有逆（T*）-1，则T*"=（T*）-1=（T1）*。Proo/. I.对于固定的y E~，Ly（x）=（T-1x，y）是定义于xE~）T-I=~ T的线性泛函.设~）Q是所有y的集合，使得Ly（x）是有界泛函.若y E）Q，则Ly（x）可推广到所有的x ET，且存在唯一的y 'E ~ T使得L~（x）=（x，y'）.我们通过Qy = y '定义一个算子Q。Q显然是一个线性算子，其定义域为~）Q，值域包含在~ T中。因此，我们有（1）对于XE~ T，y（~）Q和HQYli= LiLy] t，iv（x）=（T-lx，y）=（x，Qy），其中LiLy] t是泛函Ly与域91 T的界。二.设（1）中的T-ix= z，x= Tz，则对每个zE~）~，和yE~）Q，得到（z，y）：=（Tz，Qy）.因此，我们有QyE~ T*，T* Qy= y。这意味着yET *，因此~）Q~~ T*，~ Q~ _）T* 和T* Q= I in~）Q。另一方面，如果y= T* z，对于某个z E~）T*，则Ly（x）=（T-1 x，y）=（T-1 x，T* z）=（TT-1 x，z）=（x，z）是一个有界泛函，其界为~ t zll，因此我们有y E）Q。因此，9~ T* _ Q。从而证明了~）Q=~ T*，~ Qg ~ T*，T* Q= I in~ T*，即Q是T* 的右逆，且满足引理中的条件i），ii），iii）。三.下面我们将证明满足这些条件的算子T ′必与Q重合。通过i），T ′和Q具有相同的域~ T*，iii）证明了T*（T ′-Q）= 0。因此，对于y ET*，z ET，（T z，（T *'-Q）y）=（z，T*（T *'-Q）y）= 0，即，（T *'-Q）y与~，正交。但是，由于（T ′-Q）y ET由ii），我们必须有（T ′-Q）y= 0或T ′ = Q。四.当（T*）-1存在时，我们将其应用于T* T 'y = y（yE 9 tT *），得到T' y =（T*）-1y。但由于T ′和（T*）-~具有相同的域~ T*，我们有T ′ =（T*）-~=（T-~）*（最后一个等式是众所周知的）a）。德[尼申]岛设S和T是两个线性算子。我们将写S<< T，如果~ z~~）T和[iS xlI g liT xlI for all x E）~-。
Lemma I. Let T be a linear operator and let T* and T-~ 1 exist. Then T* has a unique right inverse T*'with the following properties2): i)~ T*"=~"*, ii)~ T*'G~ T, iii)~ T*'~) T* and T* T*'= I inST.. If in particular T* has an inverse (T*)-1, then T*"=(T*)-1=(Tl)*. Proo/. I. For a fixed y E~, Ly (x)=(T-1 x, y) is a linear functional defined for xE~) T-I=~ T. Let~) Q be the set of all y such that Ly (x) is a bounded functional. If y E) Q, Ly (x) can be extended to all x ET preserving the bound, and there is a unique y'E~ T such that L~(x)=(x, y'). We define an operator Q by Q y= y'. Q is clearly a linear operator with the domain~) Q just defined and with range contained in~ T. Thus we have (1) iv (x)=(T-Ix, y)=(x, Qy) for XE~ T, y (~) Q and HQYli= liLy] l, where liLy] t is the bound of tile functional Ly with domain 9lT. II. Setting T-ix= z, x= Tz in (1), we obtain (z, y):=(T z, Qy) for every zE~)~, and yE~) Q. Hence we have QyE~ T*, T* Qy= y. This implies y ET* and hence~) Q~~ T*,~ Q~ _) T* and T* Q= I in~) Q. On the other hand if y= T* z for some z E~) T*, then Ly (x)=(T-1 x, y)=(T-1 x, T* z)=(TT-1 x, z)=(x, z) is a bounded functional with bound~ t zll, so that we have y E) Q. Hence 9~ T* _ Q. Thus we have shown that~) Q=~ T*,~ Q g~ T* and T* Q= I in~ T*, ie, that Q is a right inverse of T* and satisfies the conditions i), ii), iii) for T*'stated in the lemma. III. We shall next show that an operator T*'satisfying these conditions must coincide with Q. By i), T*'and Q have the same domain~ T* and iii) shows that T*(T*'--Q)= 0 in~ T*. Hence (T z,(T*"-Q) y)=(z, T*(T*'-Q) y)= 0 for y ET*, z ET, ie,(T*'-Q) y is orthogonal to~,. But as (T*'-Q) y ET by ii), we must have (T*'-Q) y= 0 or T*'= Q. IV. When (T*)-1 exists, we apply it to T* T*'y= y (y E 9tT*) and obtain T*'y=(T*)-1 y. But as T*'and (T*)-~ have the same domain~ T*, we have T*'=(T*)-~=(T-~)*(The last equality is well known) a). De [inition I. Let S and T be two linear operators. We shall write S<< T if~ z~~) T and [iS xlI g liT xIl for all x E)~-.