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.


1 Answer 1


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.


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