The Triangle Inequality for Inner Product Spaces

We will now look at a very important theorem known as the triangle inequality for inner product spaces. Suppose that $V$ i an inner product space. If we form a triangle with the vectors $u$, $v$, and $u + v$, then the shortest path from the initial point of $u$ to the terminal point of $v$ is the norm $\| u + v \|$ as summarized in the following diagram:

Thus for any vectors $u, v \in V$ we have that $\| u + v \| ≤ \| u \| + \| v \|$.

Theorem 1 (The Triangle Inequality for Inner Product Spaces): Let $V$ be an inner product space with $u, v \in V$. Then: a) $\| u + v \| ≤ \| u \| + \| v \|$. b) $\| u + v \| = \| u \| + \| v \|$ is and only if $u$ or $v$ is a nonnegative scalar multiple of the other.

• Proof of a): Let $V$ be an inner product space and let $u, v \in V$. Recall that if $z \in \mathbb{C}$. Then $z = a + bi$ and $\bar{z} = a - bi$ so $z + \overline{z} = 2a = 2 \mathrm{Re} (z)$. We will use this in the proof below:

(1)

\begin{align} \quad \| u + v \|^2 = <u + v, u + v> \\ \quad \| u + v \|^2 = <u, u> + <u, v> + <v, u> + <v, v> \\ \quad \| u + v \|^2 = \| u \|^2 + \| v \|^2 + <u, v> + \overline{<u, v>} \\ \quad \| u + v \|^2 = \| u \|^2 + \| v \|^2 + 2 \mathrm{Re} (<u, v>) \\ \quad \| u + v \|^2 ≤ \| u \|^2 + \| v \|^2 + 2 \mid <u, v> \mid \\ \end{align}

• Now we will apply the Cauchy-Schwarz inequality to get that:

(2)

\begin{align} \quad \| u + v \|^2 ≤ \| u \|^2 + \| v \|^2 + 2 \| u \| \| v \| \\ \quad \| u + v \|^2 ≤ (\| u \| + \| v \|)^2 \end{align}

• We now take the square roots of both sides to get that $\| u + v \| ≤ \| u \| + \| v \|$. $\blacksquare$

• Proof of b): $\Rightarrow$ Suppose that $\| u + v \| = \| u \| + \| v \|$. Then from the proof above we must have the following equality:

(3)

\begin{align} \quad 2 \mid <u, v> \mid = 2 \| u \| \| v \| \\ \quad \mid <u, v> \mid = \| u \| \| v \| \end{align}

• We have that the use of the Cauchy-Schwarz inequality implies that then $u$ or $v$ must be a scalar multiple of the other, more precisely, a nonnegative scalar multiple.

• $\Leftarrow$ Suppose that one of $u$ or $v$ is a nonnegative scalar multiple of the other. Suppose $u = kv$ for $k \in \mathbb{F}$ ($\mathbb{R}$ or $\mathbb{C}$). Then:

(4)

\begin{align} \quad \mid <u, v> \mid = \mid <kv, v> \mid = \mid k \mid \| v \|^2 = \mid k \mid \| v \| \| v \| = \| kv \| \| v \| = \| u \| \| v \| \end{align}

• Since we have the equality above, we can see that from the proof of (a) that $\| u + v \| = \| u \| + \| v \|$. $\blacksquare$