The Riesz-Fischer Theorem - Square Lebesgue Integrable Functions

# The Riesz-Fischer Theorem for the Inner Product Space of Square Lebesgue Integrable Functions

We are now ready to present an extremely important theorem called the Riesz-Fischer theorem which will tell us that the inner product space of square Lebesgue integrable functions is complete - that is, every Cauchy sequence of functions if $L^2(I)$ converges to function in $L^2(I)$.

 Theorem 1 (The Riesz-Fischer Theorem): Let $L^2(I)$ denote the inner product space of all square-Lebesgue integrable functions on an interval $I$ with inner product $(\cdot, \cdot)$ defined for all $f, g \in L^2(I)$ by $\displaystyle{(f, g) = \int_I f(x)g(x) \: dx}$ and norm $\| f \| = (f, f)^{1/2}$ (the L2 norm). If $(f_n(x))_{n=1}^{\infty}$ is a sequence of square Lebesgue integrable functions on $I$ that is Cauchy then $(f_n(x))_{n=1}^{\infty}$ converges to a function $f$ that is Lebesgue integrable on $I$.

The Riesz-Fischer theorem can be restated as follows: “The inner product space of square-Lebesgue integrable functions on an interval $I$, $L^2(I)$, is complete.”

• Proof: Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of square Lebesgue integrable functions on $I$ that is Cauchy. Then for $\epsilon_k = \frac{1}{2^k} > 0$ there exists an $N_k \in \mathbb{N}$ such that for all $m, n \geq N_k$ we have:
(1)
\begin{align} \quad \| f_m(x) - f_n(x) \| \leq \epsilon_k = \frac{1}{2^k} \end{align}
• Let $n_1 = N_1$ and for each $k \in \{ 2, 3, … \}$ let $n_k$ be chosen such that $n_k \geq N_k$ and $n_k \geq n_{k-1}$. Then $(n_k)_{k=1}^{\infty}$ is an increasing sequence of natural numbers such that for each $k \in \{ 1, 2, … \}$ we have that:
(2)
\begin{align} \| f_{n_k}(x) - f_{n_{k+1}}(x) \| < \frac{1}{2^k} \end{align}
• Taking the sums from $k = 1$ to $\infty$ show us that:
(3)
\begin{align} \quad \sum_{k=1}^{\infty} \| f_{n_k}(x) - f_{n_{k+1}}(x) \| < \sum_{k=1}^{\infty} \frac{1}{2^k} = 1 \end{align}
• Therefore by comparison, $\displaystyle{\sum_{k=1}^{\infty} \| f_{n_k}(x) - f_{n_{k+1}}(x) \|}$ converges and so the following series of functions also converges to some function $f$ that is square Lebesgue integrable on $I$, i.e.:
(4)
\begin{align} f_{n_1}(x) + \sum_{k=1}^{\infty} [f_{n_{k+1}}(x) - f_{n_k}(x)] = f(x) \end{align}
• Now for all $m \in \mathbb{R}$ we have that:
(5)
\begin{align} \quad f(x) &= f_{n_1}(x) + \sum_{k=1}^{\infty} [f_{n_{k+1}}(x) - f_{n_k}(x)] \\ \quad f(x) &= f_{n_1}(x) + \sum_{k=1}^{m-1} [f_{n_{k+1}}(x) - f_{n_k}(x)] + \sum_{k=m}^{\infty} [f_{n_{k+1}}(x) - f_{n_k}(x) \\ \quad f(x) &= f_{n_1}(x) + [f_{n_m}(x) - f_{n_1}(x)] + \sum_{k=m}^{\infty} [f_{n_{k+1}}(x) - f_{n_k}(x) \\ \quad f(x) &= f_{n_m}(x) + \sum_{k=m}^{\infty} [f_{n_{k+1}}(x) - f_{n_k}(x) \\ \quad f(x) - f_{n_m}(x) &= \sum_{k=m}^{\infty} [f_{n_{k+1}}(x) - f_{n_k}(x) \end{align}
• Taking the norm of both sides gives us that:
(6)
\begin{align} \quad \| f(x) - f_{n_m}(x) \| = \biggr \| \sum_{k=m}^{\infty} [f_{n_{k+1}}(x) - f_{n_k}(x) \biggr \| \leq \sum_{k=m}^{\infty} \| f_{n_{k+1}}(x) - f_{n_k}(x) \| < \sum_{k=m}^{\infty} \frac{1}{2^k} = \frac{1}{2^{m-1}} \end{align}
• Taking the limit as $m \to \infty$ shows us that $(f_{n_k})_{k=1}^{\infty}$ converges to $f(x)$.
• Now since $(f_n(x))_{n=1}^{\infty}$ is a Cauchy sequence we have that for $\epsilon_1 = \frac{\epsilon}{2} > 0$ there exists an $N_1 \in \mathbb{N}$ such that for all $m, n \geq N_1$ we have that:
(7)
\begin{align} \quad \| f_m(x) - f_n(x) \| < \epsilon_1 = \frac{\epsilon}{2} \quad (*) \end{align}
• And since the subsequence $(f_{n_k}(x))_{k=1}^{\infty}$ converges to $f(x)$ for $\epsilon_1 = \frac{\epsilon}{2} > 0$ there exists an $k_1 \in \mathbb{N}$ such that for all $k \geq k_1$ we have that:
(8)
\begin{align} \quad \| f_{n_k}(x) - f(x) \| < \epsilon_1 = \frac{\epsilon}{2} \quad (**) \end{align}
• Then for all $k$ such that $n_k \geq N$ and $k \geq k_1$ we have that $(*)$ and $(**)$ hold simultaneously and so:
(9)
\begin{align} \quad \| f_n(x) - f(x) \| = \| f_n(x) - f_{n_k}(x) + f_{n_k}(x) - f(x) \| \leq \| f_n(x) - f_{n_k}(x) \| + \| f_{n_k}(x) - f(x) \| < \epsilon_1 + \epsilon+1 = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}
• Therefore the sequence $(f_n(x))_{n=1}^{\infty}$ converges to $f$ which is a square Lebesgue integrable function on $I$. $\blacksquare$