Convergence of Legendre Expansion in the Mean-Square sense.

Question

I need to show that the Legendre Expansion converges to the function in the mean-square sense, here are some definitions.

$$L_n(x)=\frac{d^n}{dx^n}(x^2-1)^n \text{ and } \mathfrak{L}_n(x)=\frac{L_n(x)}{||L_n(x)||} \text{ on } [-1,1].$$

Where the norm is defined as (as well as the specific value the above produces),

$$||L_n(x)||^2=\int_{-1}^{1}|L_n(x)|^2 \ d\theta=\frac{(n!)^22^{2n+1}}{2n+1}.$$

Additionally define an inner product,

$$\langle f,g \rangle=\int_{-1}^{1}f(x)\overline{g(x)} \ dx.$$

Where $\overline{a}$ denotes the complex conjugate. Then the Legendre Expansion is as follows,

$$\sum_{n=0}^{\infty}\langle f,\mathfrak{L}_n \rangle\mathfrak{L}_n(x)$$

Where $f$ is assumed to be Riemann integralable on $[-1,1]$. Therefore I need to show that this converges to $f$ in the mean square sense, i.e,

$$\int_{-1}^{1}\left|f(\theta)-\sum_{n=0}^{N}\langle f,\mathfrak{L}_n \rangle\mathfrak{L}_n(\theta)\right|^2 \ d\theta \to 0 \text { as } N \to \infty$$

What I've tried is using the Weirstrass Approximation Theorem, I can find a polynomial $P(x)$ such that $|f(x)-P(x)|<\epsilon$. Therefore we can rewrite the above into the following, using this $P(x)$ and the triangle inequality.

$$\int_{-1}^{1}\left|f(\theta)-\sum_{n=0}^{N}\langle f,\mathfrak{L}_n \rangle\mathfrak{L}_n(\theta)\right|^2 \ d\theta$$ $$\leq \int_{-1}^{1}\left(|f(\theta)-P(\theta)|+\left|P(\theta)-\sum_{n=0}^{N}\langle f,\mathfrak{L}_n \rangle\mathfrak{L}_n(\theta)\right|\right)^2d\theta$$

The first term is easily controlled as we defined it as such. The second term is what I am having trouble with. Any help on this would be appreciated. For clarity purposes I will include the full question here. I have proven (a) through (c) and currently working on (d). $$$$

Note that we need not show that Gram-Schimdt process.

Answer 1

Answering my own question here, so it may be false.

Define,

$$S_N(f)=\sum_{n=0}^{N}\langle f, \mathfrak{L_n}\rangle\mathfrak{L}_n(\theta)$$

Where $f$ is assumed to be Riemann integralable on $[-1,1]$ and $\mathfrak{L}_n(\theta)$ is defined in the question.

Now consider the following inner product for $m \leq N$,

$$\langle f-S_N(f),\mathfrak{L}_m \rangle=\langle f, \mathfrak{L}_m \rangle-\langle S_N(f),\mathfrak{L}_m \rangle=\langle f, \mathfrak{L}_m \rangle-\langle \sum_{n=0}^{N}\langle f, \mathfrak{L_n}\rangle\mathfrak{L}_n(\theta),\mathfrak{L}_m \rangle$$

We focus on the second term,

$$\langle \sum_{n=0}^{N}\langle f, \mathfrak{L_n}\rangle\mathfrak{L}_n(\theta),\mathfrak{L}_m \rangle=\sum_{n=0}^{N}\langle f,\mathfrak{L}_n\rangle\langle\mathfrak{L}_n(\theta),\mathfrak{L}_m\rangle$$

Since by Part A, $\langle\mathfrak{L}_n,\mathfrak{L}_m\rangle=0$ for all $m\neq n$. It remains to see its value when they are equal, namely,

$$\int_{-1}^{1}|\mathfrak{L}_n(\theta)|^2 d\theta=\frac{1}{||L_n||^2}\int_{-1}^{1}|L_n(\theta)|^2d\theta$$

Which by Part B, we calculated these two quantities to be equal so in fact it is equal to $1$, therefore we obtain that,

$$\langle f-S_N(f),\mathfrak{L}_m \rangle=\langle f, \mathfrak{L_m}\rangle-\langle f,\mathfrak{L}_m\rangle\langle\mathfrak{L}_m,\mathfrak{L}_m\rangle=0$$

So from this it follows that,

$$\langle f-S_N(f),\sum_{m=0}^{N}a_m\mathfrak{L}_m \rangle=0$$

For any coefficients $a_m$, since any linear combination of orthogonal elements are still orthogonal. Then by Pythagoras we get that,

$$||f-\sum_{m=0}^{N}a_m\mathfrak{L}_m||^2=||f-S_N(f)||^2+||\sum_{m=0}^{N}(\langle f,L_m\rangle-a_m)\mathfrak{L}_m||^2$$

Which then implies that,

\begin{equation} ||f-\sum_{m=0}^{N}a_m\mathfrak{L}_m||\geq||f-S_N(f)||. \end{equation}

Finally, by the Weierstrass Approximation Theorem, we know that for any given $f$ we can find a polynomial $P$ of degree sufficiently large, say $N$, such that $|f(x)-P(x)|<\epsilon$. Furthermore by Part C, we know that we can write $P(x)$ as a linear combination of $\{\mathfrak{L}_1,\cdots,\mathfrak{L}_N\}$, and take the coefficients to be $a_i$, $i=0,1,2,3,\cdots,N$.

Finally from earlier we get that,

$$||f-S_N(f)||\leq||f-\sum_{i=0}^{N}a_i\mathfrak{L_i}||<\epsilon.$$

Hence every Riemann Integralable $f$ on $[-1,1]$ has a Legendre Expansion.

Answer 2

Let $\{ P_n \}_{n=0}^{\infty}$ denote the Legendre polynomials, normalized so that $\|P_n\|_{L^2[-1,1]}=1$ for all $n$.

Let $f \in L^2[-1,1]$ and let $\epsilon > 0$ be given. Then there is a continuous function $g \in C[-1,1]$ such that $\|f-g\|_{L^2} < \epsilon/2$. The Weierstrass approximation theorem gives the existence of a polynomial $p$ such that $\|g-p\|_{L^2} \le \sqrt{2}\|g-p\|_{C[-1,1]} < \epsilon/2$. Therefore, there is a polynomial $p$ such that $$ \|f-p\|_{L^2} < \epsilon. $$ Hence, for any $N$ at least as large as the order of the polynomial $p$, one has $$ \left\|f - \sum_{n=0}^{N}\langle f,P_n\rangle P_n\right\| < \epsilon. $$ This is because the truncated Legendre expansion is the closest mean-square approximation to $f$ by polynomials of the given order. It follows that $\sum_{n=0}^{\infty}\langle f,P_n\rangle P_n$ converges to $f$ in $L^2[-1,1]$.