Examples of Norms

Different Spaces, Different Notions of Size

The norm axioms from Norm Axioms do not prescribe a single formula. They describe the properties a size function must satisfy, while the choice of formula depends on the vector space and on what aspects of its elements we want to measure. In a coordinate space, a formula can measure the sizes of coordinates. In a function space, it can measure values across an interval, either through a supremum or through an integral.

We will verify several examples rather than assume that a familiar-looking formula is automatically a norm. The key issue for integral formulas is positive definiteness: an integral must detect every nonzero function in the space. Continuity makes that possible, because a continuous function that is nonzero at one point must remain bounded away from zero on a small interval.

A Norm on a Polynomial Space

Let $\mathcal{P}_2$ be the real vector space of polynomials of degree at most two. Every $p\in\mathcal{P}_2$ has a unique representation $p(x)=a_0+a_1x+a_2x^2$. We can define size by assigning different positive weights to its coefficients.

Proposition: For $p(x)=a_0+a_1x+a_2x^2$, define $$ N(p)=|a_0|+2|a_1|+|a_2|. $$ Then $N$ is a norm on $\mathcal{P}_2$.

Proof. Each term in $N(p)$ is nonnegative. Their sum is zero exactly when $a_0=a_1=a_2=0$, which, by uniqueness of the coefficient representation, is exactly when $p$ is the zero polynomial. Thus $N$ is positive definite.

For a real scalar $c$, the coefficients of $cp$ are $ca_0,ca_1,ca_2$. Therefore $$ N(cp)=|ca_0|+2|ca_1|+|ca_2| =|c|\bigl(|a_0|+2|a_1|+|a_2|\bigr) =|c|N(p). $$ Finally, if $q(x)=b_0+b_1x+b_2x^2$, then the coefficients of $p+q$ are $a_0+b_0,a_1+b_1,a_2+b_2$. The absolute-value triangle inequality gives $$ N(p+q) =|a_0+b_0|+2|a_1+b_1|+|a_2+b_2| \leq N(p)+N(q). $$ All three norm axioms hold. $\square$

Worked Example: Measuring a Polynomial by Its Coefficients

For $p(x)=1-2x+3x^2$, the coefficient norm is $$ N(p)=|1|+2|-2|+|3|=1+4+3=8. $$ For $q(x)=-2+x+x^2$, $$ N(q)=|-2|+2|1|+|1|=2+2+1=5. $$ Their sum is $p+q=-1-x+4x^2$, so $$ N(p+q)=|-1|+2|-1|+|4|=1+2+4=7\leq 8+5=N(p)+N(q). $$ The weight on the linear coefficient changes how much that coefficient contributes to size, but it does not interfere with the norm axioms because the weight is positive and fixed.

The Integral of the Absolute Value

Now let $a<b$, and consider the real vector space $C[a,b]$ of continuous functions on $[a,b]$. For $f\in C[a,b]$, define its integral size by $$ \|f\|_1=\int_a^b |f(x)|\,dx. $$ The integral exists because $|f|$ is continuous. Nonnegativity and homogeneity follow directly from properties of absolute value and the integral. Positive definiteness needs the continuity of $f$: if $f(x_0)\neq0$, then $|f|$ stays positive on an interval of positive length around $x_0$, or on a one-sided interval if $x_0$ is an endpoint.

Theorem: On $C[a,b]$, the formula $$ \|f\|_1=\int_a^b |f(x)|\,dx $$ defines a norm.

Proof. Since $|f(x)|\geq0$ for every $x$, its integral is nonnegative. If $f$ is the zero function, then $\|f\|_1=0$. Conversely, suppose $f$ is not the zero function. There is an $x_0\in[a,b]$ with $|f(x_0)|>0$. Set $c=|f(x_0)|/2>0$. By continuity, there is a relative neighborhood of $x_0$ in $[a,b]$ on which $|f(x)|>c$. This neighborhood contains an interval of positive length, including when $x_0=a$ or $x_0=b$, because $a<b$. Thus $|f|$ has a positive lower bound on an interval of positive length, and consequently $\int_a^b|f(x)|\,dx>0$. This proves positive definiteness.

For every real $r$, $|rf(x)|=|r||f(x)|$, so linearity of the integral gives $$ \|rf\|_1=\int_a^b|rf(x)|\,dx =|r|\int_a^b|f(x)|\,dx =|r|\|f\|_1. $$ For $f,g\in C[a,b]$, the pointwise inequality $|f(x)+g(x)|\leq|f(x)|+|g(x)|$, together with monotonicity and additivity of the integral, gives $$ \|f+g\|_1 =\int_a^b|f(x)+g(x)|\,dx \leq\int_a^b|f(x)|\,dx+\int_a^b|g(x)|\,dx =\|f\|_1+\|g\|_1. $$ The three axioms hold, so this formula is a norm. $\square$

Worked Example: An Integral Norm of a Continuous Function

On $[0,1]$, take $f(x)=x^2$. It is nonnegative, so its integral norm is $$ \|f\|_1=\int_0^1 x^2\,dx =\left[\frac{x^3}{3}\right]_0^1 =\frac{1}{3}. $$ For the constant function $g(x)=-2$, the absolute value is $2$ throughout the interval, giving $$ \|g\|_1=\int_0^1 2\,dx=2. $$ The sign of a function does not cancel contributions to this norm, because the integrand is its absolute value.

The Quadratic-Integral Norm

A second useful formula on $C[a,b]$ is $$ \|f\|_2=\left(\int_a^b |f(x)|^2\,dx\right)^{1/2}. $$ As before, nonnegativity, homogeneity, and positive definiteness are natural to check. The triangle inequality is less immediate: it does not follow just by applying the pointwise triangle inequality and integrating. Instead, we first establish an integral form of the Cauchy–Schwarz inequality.

Theorem (Integral Cauchy–Schwarz Inequality): For $f,g\in C[a,b]$, $$ \left|\int_a^b f(x)g(x)\,dx\right| \leq \left(\int_a^b f(x)^2\,dx\right)^{1/2} \left(\int_a^b g(x)^2\,dx\right)^{1/2}. $$

Proof. If $\int_a^b g(x)^2\,dx=0$, continuity and nonnegativity imply that $g$ is identically zero, by the positive-definiteness argument used for the integral norm. The left side is then zero, and the inequality holds.

Otherwise, write $A=\int_a^b g(x)^2\,dx>0$, $B=\int_a^b f(x)g(x)\,dx$, and $C=\int_a^b f(x)^2\,dx$. For every real $t$, the square $(f(x)-tg(x))^2$ is nonnegative, so $$ 0\leq\int_a^b(f(x)-tg(x))^2\,dx =C-2tB+t^2A. $$ Choose $t=B/A$. Substitution yields $$ 0\leq C-\frac{B^2}{A}, $$ and hence $B^2\leq AC$. Taking nonnegative square roots proves the stated inequality. $\square$

Theorem: On $C[a,b]$, the formula $$ \|f\|_2=\left(\int_a^b f(x)^2\,dx\right)^{1/2} $$ defines a norm.

Proof. The integral of $f^2$ is nonnegative, so its square root is nonnegative. It is zero exactly when $\int_a^b f^2=0$, which, since $f$ is continuous, happens exactly when $f$ is identically zero. For any real $r$, $$ \|rf\|_2 =\left(\int_a^b r^2f(x)^2\,dx\right)^{1/2} =|r|\|f\|_2. $$ It remains to verify the triangle inequality. Expanding the square and applying the Integral Cauchy–Schwarz Inequality gives $$ \begin{aligned} \|f+g\|_2^2 &=\int_a^b(f(x)+g(x))^2\,dx\\ &=\|f\|_2^2+2\int_a^b f(x)g(x)\,dx+\|g\|_2^2\\ &\leq\|f\|_2^2+2\|f\|_2\|g\|_2+\|g\|_2^2\\ &=(\|f\|_2+\|g\|_2)^2. \end{aligned} $$ Both sides before squaring are nonnegative, so taking square roots gives $\|f+g\|_2\leq\|f\|_2+\|g\|_2$. Thus $\|\cdot\|_2$ satisfies all three axioms. $\square$

Worked Example: The Quadratic-Integral Norm

On $[0,1]$, let $f(x)=1+x$. Since $f(x)^2=1+2x+x^2$, $$ \|f\|_2 =\left(\int_0^1(1+2x+x^2)\,dx\right)^{1/2} =\left(1+1+\frac{1}{3}\right)^{1/2} =\sqrt{\frac{7}{3}}. $$ For the constant function $g(x)=2$, $$ \|g\|_2=\left(\int_0^1 4\,dx\right)^{1/2}=2. $$ These calculations illustrate that the quadratic-integral norm squares the function before integrating, then takes a square root; it is not the same formula as $\|\cdot\|_1$.

What an Integral Norm Detects

The supremum norm on bounded functions, established earlier in The Supremum Norm, measures the largest value a function takes or approaches. Integral norms behave differently: they accumulate the size of a function across the interval. For example, on $[0,1]$, define $$ f_n(x)=\max\bigl(1-n|x-\tfrac12|,0\bigr),\qquad n\geq2. $$ Each $f_n$ is continuous, has supremum $1$, and forms a triangle with base length $2/n$ and height $1$. Therefore $$ \|f_n\|_\infty=1,\qquad \|f_n\|_1=\frac{1}{n}. $$ The integral norm becomes small because the part of the interval where $f_n$ is nonzero becomes narrow, even though the maximum value remains $1$. This is one reason that different norms can give genuinely different information about functions.

There is also an important limitation. The formula $\int_a^b|f(x)|\,dx$ is not a norm on every space of functions. For instance, on all bounded functions on $[0,1]$, the function that is $1$ at $x=1/2$ and $0$ elsewhere is nonzero but has integral of its absolute value equal to zero. It is not continuous, so the positive-definiteness proof for $C[a,b]$ does not apply. When choosing an integral formula, both the formula and the underlying vector space matter.

Takeaway: Weighted coefficient formulas and integral formulas can define norms, but each axiom must be checked on the chosen space. Continuity ensures that a nonzero continuous function is detected by the integral, while the Integral Cauchy–Schwarz Inequality supplies the triangle inequality for the quadratic-integral norm.

Check Your Understanding

Use the definitions and proofs in this tutorial to answer the following questions.

Why must all three coefficient weights be positive for the weighted polynomial formula to be positive definite?
Where does continuity enter the proof that $\int_a^b|f(x)|\,dx=0$ implies $f=0$ on $[a,b]$?
In the proof of the Integral Cauchy–Schwarz Inequality, why is it necessary to treat $\int_a^b g(x)^2\,dx=0$ separately?
Which inequality supplies the key estimate in the proof of the triangle inequality for $\|\cdot\|_2$?
Why does the function supported at a single point show that the integral formula is not a norm on all bounded functions?

What You'll Learn