Integral Norms Of Functions

From an Integral Inner Product to a Function Norm

A norm assigns a nonnegative length to each vector. In a function space, the vectors are functions, and an integral can combine the values of a function across an interval to produce one number. The resulting norm depends on both the function and the interval: changing the interval can change the amount of the function that contributes to its length.

Let \(a,b\in\mathbb{R}\) with \(a<b\), and consider the real vector space \(C([a,b],\mathbb{R})\) of continuous real-valued functions on \([a,b]\). For \(f,g\in C([a,b],\mathbb{R})\), the integral inner product is \(\langle f,g\rangle=\int_a^b f(x)g(x)\,dx\). The induced norm is the nonnegative square root of the inner product of a function with itself.

The square inside the integral is essential. Even when \(f\) takes negative values, \(f(x)^2\geq 0\) at every point, so the integral cannot cancel positive contributions against negative ones. This norm is not generally the same as integrating \(f\) itself, nor is it the same as taking the square root of the integral of \(|f|\).

The requirement that \(a<b\) matters. A nondegenerate interval has positive length, allowing a continuous function that is nonzero at some point to contribute a strictly positive amount to the integral. The next result makes that relationship precise.

Proof. If \(f(x)=0\) for every \(x\in[a,b]\), then \(f(x)^2=0\) throughout the interval, so its integral is \(0\). For the converse, suppose \(f(x_0)\ne0\) at some \(x_0\in[a,b]\). By continuity, there is a positive \(\delta\) such that \(|f(x)-f(x_0)|<|f(x_0)|/2\) whenever \(x\in[a,b]\) and \(|x-x_0|<\delta\). The triangle inequality then gives \(|f(x)|>|f(x_0)|/2\) on that part of the interval. Because \(a<b\), this part contains an interval of positive length, including when \(x_0\) is an endpoint. Thus \(f(x)^2>|f(x_0)|^2/4\) on an interval of positive length, and \(f(x)^2\geq0\) elsewhere. It follows that \(\int_a^b f(x)^2\,dx>0\), contradicting the assumed zero integral. Therefore \(f\) must be zero everywhere. \(\square\)

Computing a Polynomial Norm

To compute an integral norm, first square the function, then integrate that squared expression over the stated interval, and finally take the nonnegative square root. Expanding before integrating often makes the calculation straightforward. Keep the interval visible throughout: it determines the endpoints at which the antiderivative is evaluated.

The output is a nonnegative real number, even if a function changes sign. For example, the norm of \(x\) on a symmetric interval is positive, although the integral of \(x\) on that interval is zero. That distinction follows directly from the square in the definition.

The Interval Is Part of the Calculation

The notation \(\|f\|_{[a,b]}\) emphasizes that a function’s norm is calculated over a particular domain. Restricting the interval changes the integral, so the length assigned to the function can change too. A useful comparison follows from splitting an interval: squared norms over adjacent pieces add, because definite integrals are additive over adjacent intervals.

Proof. By the definition of the norm and the additivity of the definite integral, \(\|f\|_{[a,b]}^2=\int_a^b f(x)^2\,dx=\int_a^c f(x)^2\,dx+\int_c^b f(x)^2\,dx\). The first integral is \(\|f\|_{[a,c]}^2\), and the second is \(\|f\|_{[c,b]}^2\), again by definition. Substituting these equalities proves the formula. \(\square\)

This is an identity about squared norms, not a claim that the norm on the whole interval is the sum of the norms on the two pieces. Taking square roots gives \(\|f\|_{[a,b]}=\sqrt{\|f\|_{[a,c]}^2+\|f\|_{[c,b]}^2}\).

Trigonometric Functions and Exact Integrals

For trigonometric functions, a standard identity can turn a square into a simpler expression. In particular, \(\sin^2 x=(1-\cos 2x)/2\). Once this identity is used, the norm calculation is an ordinary definite integral. The interval remains important: evaluating the same antiderivative at different endpoints may produce different squared norms.

Common Pitfall: Integrating the Function Instead of Its Square

A common incorrect calculation is to use the integral of the function itself in place of the integral of its square. For instance, the calculation \(\sqrt{\int_0^{2\pi}\sin x\,dx}=\sqrt{0}=0\) does not compute the integral norm. It integrates \(\sin x\), whose positive and negative parts cancel, rather than \(\sin^2 x\). The correction is to follow the definition:

The positive value is consistent with the function being nonzero on the interval. More generally, for a continuous real-valued function on a nondegenerate interval, the zero-norm theorem shows that a zero norm occurs exactly for the function that is identically zero. A calculation that gives zero for a nonzero continuous function should prompt a check that the function was squared before integration.

Another useful habit is to check the interval endpoints before evaluating an antiderivative. The expression \(\int_a^b f(x)^2\,dx\) includes both the function and the specified interval; changing either one changes the quantity being computed. The interval-splitting identity provides a check when an interval is divided into pieces: the squared norm on the whole interval must equal the sum of the squared norms on those pieces.