The Triangle Inequality For Induced Norms

Why Sums Need a Norm Inequality

A norm measures the size of a vector, but the size of a sum depends on how the summands relate. Two vectors can reinforce one another or partly cancel. The triangle inequality gives a bound that works in either situation: the norm of a sum cannot exceed the sum of the separate norms. For norms induced by inner products, this fact follows by expanding a squared norm and controlling the cross term with Cauchy–Schwarz.

Let \(\mathbb{F}\) be either \(\mathbb{R}\) or \(\mathbb{C}\), and let \(V\) be an inner product space over \(\mathbb{F}\). We use the convention that the inner product is linear in its first slot and conjugate-linear in its second; over \(\mathbb{R}\), this is simply bilinearity. In either field it satisfies \(\langle \mathbf{v},\mathbf{u}\rangle=\overline{\langle \mathbf{u},\mathbf{v}\rangle}\), with conjugation doing nothing over \(\mathbb{R}\). Positive-definiteness means that \(\langle \mathbf{u},\mathbf{u}\rangle\) is real and nonnegative, and is zero exactly when \(\mathbf{u}=\mathbf{0}\).

The Cauchy–Schwarz inequality, established for real and complex inner product spaces in the earlier material, states that \(|\langle\mathbf{u},\mathbf{v}\rangle|\leq\|\mathbf{u}\|\|\mathbf{v}\|\). It applies to every pair of vectors in an inner product space; finite dimensionality is not required. The proof below uses it only to bound the cross term that appears when the squared norm of a sum is expanded.

Expanding the Squared Norm

For either field, additivity in each slot gives four terms. Hermitian symmetry pairs the two cross terms. In the complex case, their sum is twice the real part of either one; in the real case, it is twice the inner product itself.

For real inner products, \(\operatorname{Re}\langle\mathbf{u},\mathbf{v}\rangle=\langle\mathbf{u},\mathbf{v}\rangle\), so the same formula includes the real case. For a complex number \(z=a+ib\), \(\operatorname{Re}z=a\) and \(|z|=\sqrt{a^2+b^2}\), hence \(\operatorname{Re}z\leq |z|\). Combining this scalar inequality with Cauchy–Schwarz bounds the cross term above by \(2\|\mathbf{u}\|\|\mathbf{v}\|\).

Proof. For \(z=a+ib\in\mathbb{C}\), \(|z|^2=a^2+b^2\geq a^2\), so \(|\operatorname{Re}z|=|a|\leq|z|\), and therefore \(\operatorname{Re}z\leq|z|\). This also holds for real \(z\). Using the squared-norm expansion and then Cauchy–Schwarz gives

Both sides of the desired inequality are nonnegative: norms are nonnegative, and the sum of two nonnegative numbers is nonnegative. The square-root function preserves order on nonnegative real numbers. Taking square roots of the displayed inequality therefore gives \(\|\mathbf{u}+\mathbf{v}\|\leq\|\mathbf{u}\|+\|\mathbf{v}\|\), as required. \(\square\)

Worked Applications

When Does Equality Hold?

The inequality can be strict, as in the examples above, or it can be an equality. The exact condition distinguishes vectors that point in the same direction from vectors whose sum is shorter than the sum of their individual lengths. Here, “same direction” is expressed algebraically using a positive real scalar; this condition applies in complex spaces as well as real ones.

Proof. If either vector is zero, substitution gives equality directly. Now suppose both are nonzero. If \(\|\mathbf{u}+\mathbf{v}\|=\|\mathbf{u}\|+\|\mathbf{v}\|\), squaring both nonnegative sides and using the squared-norm expansion yields \(\operatorname{Re}\langle\mathbf{u},\mathbf{v}\rangle=\|\mathbf{u}\|\|\mathbf{v}\|\). Set \(t=\|\mathbf{v}\|/\|\mathbf{u}\|\), which is positive. Since \(t\) is real, expansion gives \(\|\mathbf{v}-t\mathbf{u}\|^2=\|\mathbf{v}\|^2+t^2\|\mathbf{u}\|^2-2t\operatorname{Re}\langle\mathbf{u},\mathbf{v}\rangle\). Substituting the value of \(t\) and the equality of the real part gives \(\|\mathbf{v}-t\mathbf{u}\|^2=\|\mathbf{v}\|^2+\|\mathbf{v}\|^2-2\|\mathbf{v}\|^2=0\). Positive-definiteness implies \(\mathbf{v}-t\mathbf{u}=\mathbf{0}\), so \(\mathbf{v}=t\mathbf{u}\). Conversely, suppose \(\mathbf{v}=t\mathbf{u}\) for a positive real \(t\). Then \(\mathbf{u}+\mathbf{v}=(1+t)\mathbf{u}\), and homogeneity of the induced norm gives \(\|\mathbf{u}+\mathbf{v}\|=(1+t)\|\mathbf{u}\|=\|\mathbf{u}\|+\|\mathbf{v}\|\). This proves both directions. \(\square\)

A Common Mistake

Do not replace the real part by the absolute value in the expansion. Cauchy–Schwarz gives an upper bound using the absolute value, but the expansion itself contains the real part, which can be negative. For example, take \(\mathbf{u}=(1,0)^{\top}\) and \(\mathbf{v}=(-1,0)^{\top}\) in \(\mathbb{R}^2\). A mistaken expansion using \(2|\langle\mathbf{u},\mathbf{v}\rangle|\) would give \(1+1+2|{-1}|=4\) for the squared norm of the sum. But \(\mathbf{u}+\mathbf{v}=\mathbf{0}\), so its squared norm is \(0\). The correct expansion uses \(2\langle\mathbf{u},\mathbf{v}\rangle=2(-1)\), giving \(1+1-2=0\). Only after writing the correct expansion may we bound its cross term above using \(\operatorname{Re}\langle\mathbf{u},\mathbf{v}\rangle\leq|\langle\mathbf{u},\mathbf{v}\rangle|\).

Also keep the squared and unsquared statements distinct. The expansion and Cauchy–Schwarz first show \(\|\mathbf{u}+\mathbf{v}\|^2\leq(\|\mathbf{u}\|+\|\mathbf{v}\|)^2\). The triangle inequality follows by taking square roots because both sides being compared are nonnegative.