I was going through the proof of triangle inequality as a consequence of Schwarz inequality here:


I find somethinng odd in the third step (the expansion of the inner product) . I learnt of the following properties of inner product:

  1. $\langle a,b \rangle=\langle b,a \rangle^*$

  2. $\langle a,zb+kc \rangle =z \langle a,b \rangle +k \langle a,c \rangle$

  3. $ \langle za+kb,c \rangle =z^* \langle a,c \rangle + k^* \langle b,c \rangle$

  4. $\langle za,kb \rangle = z^*k \langle a,b\rangle $

where $a,b,c$ are in general complex vectors, and $z,k$ are in general complex scalars.

But I think the third step is wrong, due to rule (1). We should have got $\langle y,x \rangle =\langle x,y \rangle^*$, and not $\langle y,x \rangle = \langle x,y \rangle$, unless both $x,y$ are taken to be real vectors. But should I think of it in this way: that the vectors involved in proving the triangle-inequality must be real, and can not at all possess an imaginary part?

  • 2
    $\begingroup$ Seems like you might be overlooking the absolute value in that third step. $\endgroup$ – Quinn Culver Dec 26 '14 at 15:59

It appears you want to know why $\langle x, y \rangle + \langle y, x \rangle \leq 2 | \langle x, y \rangle |$. This is a general property of complex numbers. Let $z = \langle x, y \rangle$. Using property $(1)$ in the definition of an inner product, your desired inequality follows from the fact that $z + z^* \leq 2 |z|$. (As you'll see from the calculation that follows, both $z + z^*$ and $2 |z|$ are real numbers, so we can compare them.) To prove this, let $z = a + bi$ with $a, b$ real. Then $z + z^* = 2a$, while $|z| = \sqrt{a^2 + b^2}$. It follows immediately that $z + z^* \leq 2 |z|$.

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  • 1
    $\begingroup$ okay.this helps. And as Quinn Culver mentioned, I did miss the abs. value sign in the third step. Thanks. $\endgroup$ – Sudeepan Datta Dec 26 '14 at 16:12
  • $\begingroup$ Glad you got it sorted out! Cheers. $\endgroup$ – Michael Joyce Dec 26 '14 at 16:14

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