Each vector space $E$ over $\mathbb{C}$ is also a vector space over $\mathbb{R}$. Show that if $(E, \langle \cdot , \cdot \rangle)$ is a complex inner product space, then

$\langle f,g \rangle_r :=$ Re$\langle f,g \rangle \quad f,g \in E$

is a real inner product on $E$ satisfying $\langle if, ig \rangle_r = \langle f,g \rangle_r$ for all $f,g \in E$.

Conversely, show that if $\langle \cdot , \cdot \rangle$ is a real inner product on the $\mathbb{C}$-vector space $E$ such that $\langle if,ig \rangle = \langle f,g \rangle$ for all $f,g \in E$, then

$\langle f,g \rangle_c := \langle f,g \rangle + i \langle f, ig \rangle$

is the unique complex inner product on $E$ with $\langle \cdot , \cdot \rangle_{cr} = \langle \cdot, \cdot \rangle$.

Based on what I know, then I would have to show that the inner product $\langle f,g \rangle_r$ satisfies the four properties in my textbook: sesquilinearity, symmetric, positive, and definite. Is this the right direction to go in order to solve this problem?

The textbook that I am using is Functional Analysis An Elementary Introduction by Markus Hasse. Any help on this problem will be greatly appreciated since I am unsure whether this is the appropriate way to handle it. Thanks in advance.

  • $\begingroup$ Remember for the complex inner product that we have conjugate symmetry. $\endgroup$
    – Mnifldz
    Feb 3, 2015 at 22:22

1 Answer 1


Yes, except when the product is real then it's not sesquilinear it's just linear in the first argument.

So for the first part you need to show:

(1) $\langle f,g \rangle_r = \langle g,f \rangle_r$

(2) For $\lambda \in \mathbb R$ we have $\lambda \langle f,g\rangle_r = \langle \lambda f,g \rangle_r$ and for $h \in E$ we have $\langle f+h,g\rangle_r = \langle f,g \rangle_r + \langle h,g\rangle_r$

(3) $\langle f,f \rangle_r \ge 0$

(4) $\langle f,f \rangle_r = 0$ if and only if $f = 0$

and in addition to this you need to show the condition $\langle if,ig \rangle_r = \langle f,g\rangle_r$.

For the second part you need to show that $\langle f,g \rangle_c$ is a complex inner product:

(1) $\langle f,g \rangle_c = \overline{\langle g,f \rangle_c}$

(2) For $\lambda \in \mathbb C$ we have $\lambda \langle f,g\rangle_c = \langle \lambda f,g \rangle_c$ and for $h \in E$ we have $\langle f+h,g\rangle_c = \langle f,g \rangle_c + \langle h,g\rangle_c$

(3) $\langle f,f \rangle_c \ge 0$

(4) $\langle f,f \rangle_c = 0$ if and only if $f = 0$

I'm a bit confused about your remaining condition since you never define $\langle f,g\rangle_{cr}$. I presume you want to show that the real part of $\langle f,g\rangle_{c}$ equals $\langle f,g \rangle$ but I might be wrong.

  • $\begingroup$ Ok, but what operations define the inner product $\langle f,g \rangle_r$? What I mean by this is I have seen some inner products involve summations and others integrals, for example. I want to say that the one for this problem involves a summation, but I am unsure. Thanks, for the advice. $\endgroup$
    – Jamil_V
    Feb 10, 2015 at 3:13
  • $\begingroup$ @Jamil_V What operations define $\langle\cdot,\cdot\rangle_r$ depends on what operations define $\langle\cdot,\cdot\rangle$ since $\langle\cdot,\cdot\rangle_r$ is defined to be the real part of $\langle\cdot,\cdot\rangle$. If $E$ is the space $L^2$ then the definition of $\langle\cdot,\cdot\rangle$ involves an integral. But since you don't have more information about $E$ you cannot say more about $\langle\cdot,\cdot\rangle_r$. $\endgroup$ Feb 11, 2015 at 23:33

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