# How do I prove that the inner product operator is a metric?

Warning: Contains erroneous definitions. See accepted answer.

I want to understand that the inner product is a positive definite metric. A metric (a.k.a distance function) is defined as a map $$d: M \times M \to \mathbb{R}$$ for some topological space $$M$$ and $$\{x,y,z\} \subset M$$, such that:

\begin{align*} d(x,y) &\geq 0 \\ d(x,y) &= 0 \Leftrightarrow x = y \\ d(x,y) &= d(y,x) \\ d(x,y) &\leq d(x,z) + d(z,y) \end{align*}

An inner product is defined for $$\{u,v,w\} \subset M$$ as an operator $$\langle \cdot,\cdot \rangle: M \times M \to \mathbb{R}$$ that satisfies:

\begin{align*} \langle v,w \rangle &\geq 0 \\ \langle v,w \rangle &= 0 \Leftrightarrow v = w = 0 \\ \langle v,w \rangle &= \langle w,v \rangle \\ \langle u+v,w \rangle &= \langle u,w \rangle + \langle v,w \rangle \\ \langle \alpha v,w \rangle &= \alpha \langle v,w \rangle \\ \end{align*}

Please correct me if the definitions are wrong. Clearly, this operator satisfies the first three conditions of a metric. It is not as obvious to me that the last two conditions, which enforce linearity of the inner product, also enforce the triangle inequality.

I have seen proofs that the Euclidean norm obeys $$\lVert x + y \rVert \leq \lVert x \rVert + \lVert y \rVert$$, but I don't think that is what I'm looking for. I understand that the inner product always induces a norm, however it seems to me that those are two distinct metrics. If the inner product is also a metric, then it should be possible to prove that using only the definitions provided above.

Using consistent notation, I want to be satisfied that:

$$\langle u,v \rangle \leq \langle u,w \rangle + \langle w,v \rangle$$

Or, equivalently, that $$\langle u,v \rangle \leq \langle u + w, v \rangle$$. I can see that the equality holds for the trivial case of $$u = v = 0$$, but I'm not sure where to go from there. Or am I completely wrong and only $$\sqrt{\langle u,v \rangle}$$ is a metric, and not $$\langle u,v \rangle$$ itself?

Your definition of inner product is wrong. Your first and last properties are contradictory: $$0\leq \langle v,w\rangle=-\langle -v,w\rangle\leq0,$$ so the only choice you have is $$\langle v,w\rangle=0$$ for all $$v,w$$.
Instead, in the definition of inner product what you require is that $$\langle v,v\rangle\geq0$$.
Your second property is also wrong, again in easy contradiction with your last property: it should be $$\langle v,v\rangle=0\implies v=0$$. Otherwise you miss the crucial feature of an inner product which is orthogonality.
Another important feature that you miss is that $$M$$ should be a vector space. Otherwise, $$u+v$$ makes no sense, nor would multiplication by a scalar.
The way to obtain a distance from an inner product, is by $$d(x,y)=\sqrt{\langle x-y,x-y\rangle}.$$ This is usually phrased in terms of a norm, i.e. $$d(x,y)=\|x-y\|,$$ where the norm is $$\|x\|=\sqrt{\langle x,x\rangle}.$$