Let $\langle x,y\rangle$ be the scalar product of $x$ and $y$ in a linear space $X$ over either $\mathbb{R}$ or $\mathbb{C}$. This scalar product satisfies the three properties: Bilinerity/Sesquilinearity, Symmetry/Skew symmetry, and Positivity.

The Schwarz inequality for this scalar product is: $$ \lvert\langle x,y\rangle\rvert \leq \lVert x \rVert \lVert y\rVert \tag{*} $$ where $\lVert x \rVert = \langle x,x\rangle^{1/2}$, the induced norm of $x$.

The following proof for this theorem is taken from Lax's functional analysis book: Let $t \in \mathbb{R}$ and $y \in X \neq 0$, then: $$ \lVert x+ty \rVert^2=\lVert x \rVert^2 + 2tRe(\langle x,y\rangle) + t^2 \lVert y \rVert ^2 \tag{1} $$ Set: $$ t = -Re(\langle x,y\rangle)/\lVert y \rVert ^2 $$ and multiply (1) by $\lVert y \rVert^2 $, we get: $$ Re^2(\langle x,y\rangle) \leq \lVert x \rVert^2\lVert y \rVert^2 $$ Replacing $x = ax$, $\lvert a \rvert =1$, so chosen that $a\langle x,y\rangle\in \mathbb{R}$, we deduce (*). The equality holds in (*) iff $x$ and $y$ are scalar multiples of one another.

The answer to any of these questions would be appreciated:

  1. What is the motivation behind considering the quantity $\lVert x +ty \rVert^2$ in (1)?

  2. Similarly, what is the intuition behind setting $t=-Re(\langle x,y\rangle)/\lVert y \rVert^2$?

  3. Finally, what is the intuition behind setting $x=ax, \lvert a \rvert =1$? Moreover, how do we deduce (*) from setting $x=ax, |a|=1$, and $a$ chosen such that $a\langle x,y\rangle$ real? How do we even know that such a decomposition of $x$ into $ax$ is always possible?

  • 1
    $\begingroup$ I am not sure there is any answer other than it works. $\endgroup$ Mar 19, 2019 at 23:38

2 Answers 2


The short answer to each of your questions is that we're deliberately customising our observations to prove the strongest facts we can, from the simplest tricks we have.

In answer to 1, the motivation is that, whenever possible, you should try to derive an inequality from the fact that squares of certain quantities (in this case vectors) cannot be negative. Sure, this doesn't prove every known inequality, but it's got easily the best bang for your buck among the simple ideas.

As for 2, to understand the choice of $t$ it helps to first work out what happens in a vector space over $\Bbb R$. Because $\Vert x+ty\Vert^2\ge 0$ is strongest as a fact when the left-hand side is minimised, which occurs at $t=-\frac{\langle x,\,y\rangle}{\Vert y\Vert^2}$ (because minimising a real-to-real quadratic function is easy), the analogous choice of $t$ in the complex case is the one Lax uses.

Finally (for 3), choosing a unit complex number $a$ so that $\langle ax,y\rangle$ is a non-negative real number, its squared real part is just $|\langle x,\,y\rangle|^2$, which in general $\le\Re^2\langle x,\,y\rangle$ since $u,\,v\in\Bbb R\implies u^2\le u^2+v^2=|u+iv|^2$. I think Lax may have confused you a little with what he's saying here, so let me try to spell it out a bit:

  • We've proven that any vectors $X,\,y$ satisfy $\Re^2\langle x,\,y\rangle\le\Vert X\Vert^2\Vert y\Vert^2$. Yes, the capitalisation of $X$ is deliberate on my part here.
  • We write an arbitrary $x$ as $aX$ with $|a|=1$, where $\langle X,\,y\rangle\ge 0$. (You can work out what $a$ should be if you know the phase of $\langle x,\,y\rangle$, although you also need to know whether, in this convention, the inner product is linear or antilinear in its first argument, since the two options require opposite phases for $a$.) Then $$|\langle x,\,y\rangle|^2=|\langle X,\,y\rangle|^2=\Re^2\langle x,\,y\rangle\le\Vert X\Vert^2\Vert y\Vert^2=\Vert x\Vert^2\Vert y\Vert^2.$$Finally, we take the square root.

Well... I don't know if this will help, but:

  1. You consider the segment from $x$ to $y$ is a standard thing to try, as that norm "contains" also the scalar product $\langle x,y\rangle$ (or something similar to it).

  2. minimizing the norm (try to take the derivative w.r. of $t$): this is also something you often try

  3. is a standard trick with complex numbers:

    • you can always find one such $a$: $\langle x,y\rangle$ is a complex number, so it can be written as $c+id$. If you multiply it by $c-id$ what happens? Then simply normalize.
    • then the substitution is indeed tricky, written like that. Try to see it this way: we reached the equation $Re^2(\langle x,y\rangle) \leq ||x|| ||y||$ using $x$, can we repeat the same steps using $ax$? Yes. What would we have reached? Thus, instead of repeating everything, we simply substitute $x=ax$.

As a general suggestion for real analysis proofs (at least these easy ones): often using complex numbers only results in extra care to handle real/complex part, but no "true" addition. If you try to rework the proofs using only reals, you should get a quicker intuition.

In this case I suggest you recall that the scalar product is connected to the cosine of the angle between the two vectors: try to draw some 2d vectors, you might get some geometrical intuition.

  • $\begingroup$ Could you elaborate #3 a little more? In particular, how do we deduce (*) from setting $x=ax, |a|=1$, and $a$ chosen such that $a<x,y>$ real? Moreover, how do we know that such a decomposition of $x$ into $ax$ is always possible? $\endgroup$ Mar 20, 2019 at 5:00
  • $\begingroup$ Ok, I modified the original comment, now it should be more clear (and correct) $\endgroup$
    – GivAlz
    Mar 20, 2019 at 8:47

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