Let $(V,\|\cdot\|)$ be a normed space. A sequence $\{x_m\}$ is said to converge in norm to $x$ if $(*)\lim_{m \rightarrow \infty} \|x-x_m\|=0$.

Let $E$ be the set of piecewise continuous functions $[-\pi,\pi] \rightarrow \mathbb{C}$. Given the inner product $$\langle f,g \rangle :=\frac{1}{\pi} \int_{-\pi}^\pi f(x) \overline{g(x)} dx$$ we can define a norm $\| \cdot\|$ by putting $\| \cdot \| := \sqrt{\langle \cdot,\cdot \rangle}$. Explicitly, $$\|f\|^2=\langle f,f\rangle=\frac{1}{\pi} \int f(x)\overline{f(x)} \, dx = \frac{1}{\pi} \int |f(x)|^2 \, dx$$

My question relates to the following comment in my literature:

The closure property of the trigonometric orthonormal system... implies that the Fourier series of each $f \in E$ converges in norm to $f$. In other words, if the $a_n$ and $b_n$ are the Fourier coefficients of $f$, then... $$\lim_{m \rightarrow \infty} \int \left| f(x)-\left( \frac{a_0}{2}+\sum_{n=1}^m [a_n \cos{nx}+b_n \sin{nx}] \right) \right|^2 \, dx=0 \tag{**}$$

The quote speaks of convergence in norm, but the expression $(**)$ doesn't agree with the definition of convergence in norm, i.e. the expression $(*)$. With the norm defined and letting $x=f$, $\{x_m\}=\{ \frac{a_0}{2}+\sum_{n=1}^m [a_n \cos{nx}+b_n \sin{nx}] \}$, the expression $(*)$ becomes

$$\lim_{m \rightarrow \infty} \left( \frac{1}{\pi} \int \left| f(x) - \left(\frac{a_0}{2}+\sum[a_n \cos{nx}+b_n \sin{nx}] \right) \right|^2dx \right)^{1/2}=0 \tag{*$'$}$$

Again, $(*')$ and $(**)$ don't agree. I guess they do agree, but how?

If we put $x-x_m=y_m$, $a=1/\pi$, then $$(*')=\lim_{m \rightarrow \infty} \left( a \int |y_m|^2 dx \right)^{1/2}$$ The resulting constant $a^{1/2}$ doesn't really influence the result so in a sense we have $(*')=\lim \left( \int |y_m|^2dx \right)^{1/2}$, but how that equals $\lim \int |y_m|^2dx$ is what I don't get.

up vote 4 down vote accepted

The equation $(\ast\ast)$ is equivalent to norm convergence, since it is

$$\lim_{m\to\infty} \pi\cdot\lVert f - x_m\rVert^2 = 0.$$

The constant factor just multiplies the limit, but since that is $0$, it has no effect. And for a sequence of real numbers, we have $a_n \to 0$ if and only if $a_n^2 \to 0$, so saying the squares of the norms converge to $0$ and saying the norms converge to $0$ are equivalent.

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.