I tried to simplify the Cauchy - Schwarz inequality proof so that it is easier for me to remember in the long run. I took pieces from here and there and tried to minimize the length, so It'd be appreciated if anyone could verify this:

Let $V$ be an inner product space $\implies \forall u,v \in V$ holds that $\lvert \langle u,v\rangle\rvert \leq\lVert u \rVert \cdot\lVert v\lVert$ (with the equality holding iff $u,v$ are linearly dependent)

My attempt:

If either $u,v$ are equal to zero, then the above holds true.

Suppose that $u,v \ne 0$

If $u$ and $v$ are linearly dependent, which means that $u=λv$, for some $λ \in \Bbb F$, it again holds true.

Suppose that they are not linearly dependent. This means that $u_v,u_{\perp v}$ are perpendicular, (where $u_v$ is the projection of $u$ on $v$ and $u_{\perp v}=-u_v+u$). So the Pythagorean Theorem holds for $u,u_v,u_{\perp v}$ like so:

$\lVert u \rVert ^2 = \lVert u_v \rVert ^2 +\lVert u_{\perp v} \rVert ^2= \lvert \frac{\langle v,u\rangle}{\langle v,v \rangle} \rvert ^2 \cdot \lVert v\rVert^2+\lVert u_{\perp v} \rVert ^2 \ge \frac{\lvert\langle v,u\rangle \rvert ^2}{\lVert v\rVert ^2}$

taking roots and multiplying $\implies$ Cauchy Schwarz inequality

Is this proof correct?


seems alright besides the statement of when does the equality holds.

Equality holds when $u$ and $v$ are linearly dependent.

  • $\begingroup$ Oops, i will edit it $\endgroup$ – ZeroPancakes Sep 5 '17 at 19:13

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