- #1
- 19,865
- 10,850
Definition/Summary
the Schwarz inequality (also called Cauchy–Schwarz inequality and Cauchy inequality) has many applications in mathematics and physics.
For vectors a,b in an inner product space over \mathbb C:
\|a\|\|b\| \geq |(a,b)|
For two complex numbers a,b :
|a|^2|b|^2 \geq |ab|^2
In bra-ket notation, which is commonly used in Quantum Mechanics:
| \langle a | a \rangle |^2| \langle b | b \rangle |^2 \geq | \langle a | b \rangle |^2
For functions a, b which maps a real numbers into complex: x \in \mathbb{R}, a(x), \: b(x) \in \mathbb{C}:
\int |a (x)|^2 dx\int |b(x)|^2 dx \geq \left| \int a^*bdx \right|^2 ,
if the integrals have finite values.
Equations
The Schwarz inequality may be written in a number of equivalent ways:
\|a\|\|b\| \geq |(a,b)|~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)
|a|^2|b|^2 \geq |ab|^2 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)
| \langle a | a \rangle |^2| \langle b | b \rangle |^2 \geq | \langle a | b \rangle |^2 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)
\int |a (x)|^2 dx\int |b(x)|^2 dx \geq \left| \int a(x)^*b(x) \, dx \right|^2 ~~~(4)
Extended explanation
Theorem:
If a and b are vectors in an inner product space X over \mathbb C[/itex], then<br /> <br /> <div style="margin-left: 20px">|(a,b)| \leq \|a\|\|b\|​</div><br /> where the norm is the standard norm on an inner product space.<br /> <br /> <b>Proof #1:</b><br /> <br /> Let t be an arbitrary complex number.<br /> <br /> <div style="margin-left: 20px">0 \leq (a+tb,a+tb)=\|a\|^2+t(a,b)+t^*(b,a)+|t|^2\|b\|^2​</div><br /> <div style="margin-left: 20px">=\|a\|^2+2\mbox{ Re}(t(a,b))+|t|^2\|b\|^2​</div><br /> The inequality is obviously satisfied when the real part of t(a,b) is non-negative, so we can only learn something interesting when it's negative. Let's choose Arg t so that it is<br /> <br /> <div style="margin-left: 20px">=\|a\|^2-2|t||(a,b)|+|t|^2\|b\|^2​</div><br /> Now let's choose |t| so that it minimizes the sum of the last two terms. (This should give us the most interesting result).<br /> <br /> <div style="margin-left: 20px">s=|t|, A=\|b\|^2, B=2|(a,b)|​</div><br /> <div style="margin-left: 20px">f(s)=As^2-Bs​</div><br /> <div style="margin-left: 20px">f&#039;(s)=2As-B=0 \implies s=\frac{B}{2A} = \frac{|(a,b)|}{\|b\|^2}​</div><br /> <div style="margin-left: 20px">f&#039;&#039;(s)=2A&gt;0​</div><br /> Continuing with this value of |t|...<br /> <br /> <div style="margin-left: 20px">=\|a\|^2-2\frac{|(a,b)|}{\|b\|^2}|(a,b)|+\frac{|(a,b)|^2}{\|b\|^4}\|b\|^2​</div><br /> <div style="margin-left: 20px">=\|a\|^2-\frac{|(a,b)|^2}{\|b\|^2}​</div><br /> Thus, 0 \leq \|a\|^2\|b\|^2 - |(a,b)|^2<br /> <br /> <br /> <b>Proof #2:</b><br /> <br /> A proof for Ineq(3) (the last inequality in the "equations" section) will now be given, but first some basic properties of complex numbers...<br /> <br /> z = \text{Re}z + i \text{Im}z<br /> z^* = \text{Re}z - i \text{Im}z<br /> |z|^2 = z^*z \geq 0<br /> |zy|^2 = (zy)^*zy<br /> <br /> Now let c(x) = a(x) + d \, b(x), d \in \mathbb{C}.<br /> <br /> \int |c(x)|^2 dx \geq 0<br /> <br /> \int |a(x)|^2dx + d^*d \int |b(x)|^2dx + d \int a(x)^*b(x) dx + d^* \int a(x)b(x)^* dx \geq 0<br /> <br /> if the integrals have finite value, this inequality must hold for all d, we can choose:<br /> <br /> d = - \left( \int b(x)^*a(x) dx \right) / \left( \int |b(x)|^2 dx \right)<br /> <br /> \int |a(x)|^2 dx \int |b(x)|^2 dx \geq \int a(x)^*b(x)dx \int a(x)b(x)^*dx = \left| \int a(x)^*b(x)dx \right|^2<br /> <br /> * This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
the Schwarz inequality (also called Cauchy–Schwarz inequality and Cauchy inequality) has many applications in mathematics and physics.
For vectors a,b in an inner product space over \mathbb C:
\|a\|\|b\| \geq |(a,b)|
For two complex numbers a,b :
|a|^2|b|^2 \geq |ab|^2
In bra-ket notation, which is commonly used in Quantum Mechanics:
| \langle a | a \rangle |^2| \langle b | b \rangle |^2 \geq | \langle a | b \rangle |^2
For functions a, b which maps a real numbers into complex: x \in \mathbb{R}, a(x), \: b(x) \in \mathbb{C}:
\int |a (x)|^2 dx\int |b(x)|^2 dx \geq \left| \int a^*bdx \right|^2 ,
if the integrals have finite values.
Equations
The Schwarz inequality may be written in a number of equivalent ways:
\|a\|\|b\| \geq |(a,b)|~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)
|a|^2|b|^2 \geq |ab|^2 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)
| \langle a | a \rangle |^2| \langle b | b \rangle |^2 \geq | \langle a | b \rangle |^2 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)
\int |a (x)|^2 dx\int |b(x)|^2 dx \geq \left| \int a(x)^*b(x) \, dx \right|^2 ~~~(4)
Extended explanation
Theorem:
If a and b are vectors in an inner product space X over \mathbb C[/itex], then<br /> <br /> <div style="margin-left: 20px">|(a,b)| \leq \|a\|\|b\|​</div><br /> where the norm is the standard norm on an inner product space.<br /> <br /> <b>Proof #1:</b><br /> <br /> Let t be an arbitrary complex number.<br /> <br /> <div style="margin-left: 20px">0 \leq (a+tb,a+tb)=\|a\|^2+t(a,b)+t^*(b,a)+|t|^2\|b\|^2​</div><br /> <div style="margin-left: 20px">=\|a\|^2+2\mbox{ Re}(t(a,b))+|t|^2\|b\|^2​</div><br /> The inequality is obviously satisfied when the real part of t(a,b) is non-negative, so we can only learn something interesting when it's negative. Let's choose Arg t so that it is<br /> <br /> <div style="margin-left: 20px">=\|a\|^2-2|t||(a,b)|+|t|^2\|b\|^2​</div><br /> Now let's choose |t| so that it minimizes the sum of the last two terms. (This should give us the most interesting result).<br /> <br /> <div style="margin-left: 20px">s=|t|, A=\|b\|^2, B=2|(a,b)|​</div><br /> <div style="margin-left: 20px">f(s)=As^2-Bs​</div><br /> <div style="margin-left: 20px">f&#039;(s)=2As-B=0 \implies s=\frac{B}{2A} = \frac{|(a,b)|}{\|b\|^2}​</div><br /> <div style="margin-left: 20px">f&#039;&#039;(s)=2A&gt;0​</div><br /> Continuing with this value of |t|...<br /> <br /> <div style="margin-left: 20px">=\|a\|^2-2\frac{|(a,b)|}{\|b\|^2}|(a,b)|+\frac{|(a,b)|^2}{\|b\|^4}\|b\|^2​</div><br /> <div style="margin-left: 20px">=\|a\|^2-\frac{|(a,b)|^2}{\|b\|^2}​</div><br /> Thus, 0 \leq \|a\|^2\|b\|^2 - |(a,b)|^2<br /> <br /> <br /> <b>Proof #2:</b><br /> <br /> A proof for Ineq(3) (the last inequality in the "equations" section) will now be given, but first some basic properties of complex numbers...<br /> <br /> z = \text{Re}z + i \text{Im}z<br /> z^* = \text{Re}z - i \text{Im}z<br /> |z|^2 = z^*z \geq 0<br /> |zy|^2 = (zy)^*zy<br /> <br /> Now let c(x) = a(x) + d \, b(x), d \in \mathbb{C}.<br /> <br /> \int |c(x)|^2 dx \geq 0<br /> <br /> \int |a(x)|^2dx + d^*d \int |b(x)|^2dx + d \int a(x)^*b(x) dx + d^* \int a(x)b(x)^* dx \geq 0<br /> <br /> if the integrals have finite value, this inequality must hold for all d, we can choose:<br /> <br /> d = - \left( \int b(x)^*a(x) dx \right) / \left( \int |b(x)|^2 dx \right)<br /> <br /> \int |a(x)|^2 dx \int |b(x)|^2 dx \geq \int a(x)^*b(x)dx \int a(x)b(x)^*dx = \left| \int a(x)^*b(x)dx \right|^2<br /> <br /> * This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!