• Equivalently, we can show \begin{align*} {\left\lVert {x} \right\rVert}^2 - \sum_{n=1}^\infty {\left\lvert { {\left\langle {x},~{u_n} \right\rangle} } \right\rvert}^2 \geq 0 .\end{align*}

• Claim: the LHS is the norm of an element in \(H\), and thus non-negative. More precisely, set \(S_N\coloneqq\sum_{n=1}^N {\left\langle {x},~{u_n} \right\rangle}u_n\), then the above is equal to \begin{align*} {\left\lVert {x - \lim_{N\to\infty} S_N} \right\rVert}^2 .\end{align*} Note that if this is true, we’re done.

• To see this, expand the norm in terms of inner products: ` \begin{align*} {\left\lVert {x - S_N} \right\rVert}^2 &= {\left\langle {x-S_N},~{x-S_N} \right\rangle} \ &= {\left\langle {x},~{x} \right\rangle} - {\left\langle {x},~{S_N} \right\rangle} - {\left\langle {S_N},~{x} \right\rangle} + {\left\langle {S_N},~{S_N} \right\rangle} \ &= {\left\lVert {x} \right\rVert}^2 + {\left\lVert {S_N} \right\rVert}^2 - \qty{{\left\langle {x},~{S_N} \right\rangle} + {\overline{{{\left\langle {x},~{S_N} \right\rangle}}}} } \ &= {\left\lVert {x} \right\rVert}^2 + {\left\lVert {S_N} \right\rVert}^2 - 2\Re\qty{{\left\langle {x},~{S_N} \right\rangle} } \ &= {\left\lVert {x} \right\rVert}^2 + {\left\lVert {S_N} \right\rVert}^2 - 2\Re\qty{ {\left\langle {x},~{\sum_^N {\left\langle {x},~{u_n} \right\rangle} u_n } \right\rangle} } \ &= {\left\lVert {x} \right\rVert}^2 + {\left\lVert {S_N} \right\rVert}^2 - 2\Re\qty{ \sum_^N {\left\langle {x},~{{\left\langle {x},~{u_n} \right\rangle} u_n } \right\rangle} } \ &= {\left\lVert {x} \right\rVert}^2 + {\left\lVert {S_N} \right\rVert}^2 - 2\Re\qty{ \sum_^N {\overline{{{\left\langle {x},~{u_n} \right\rangle} }}} {\left\langle {x},~{u_n } \right\rangle} } \ &= {\left\lVert {x} \right\rVert}^2 + {\left\lVert {S_N} \right\rVert}^2 - 2\Re \sum_^N {\left\lvert {{\left\langle {x},~{u_n} \right\rangle} } \right\rvert}^2 \ &= {\left\lVert {x} \right\rVert}^2 + {\left\lVert {S_N} \right\rVert}^2 - 2\sum_^N {\left\lvert {{\left\langle {x},~{u_n} \right\rangle} } \right\rvert}^2 \ &= {\left\lVert {x} \right\rVert}^2 + {\left\lVert {\sum_^N {\left\langle {x},~{u_n} \right\rangle} u_n} \right\rVert}^2 - 2\sum_^N {\left\lvert {{\left\langle {x},~{u_n} \right\rangle} } \right\rvert}^2 \ &= {\left\lVert {x} \right\rVert}^2 + {\left\langle {\sum_^N {\left\langle {x},~{u_n} \right\rangle} u_n},~{\sum_^N {\left\langle {x},~{u_m} \right\rangle} u_m} \right\rangle}

  • 2\sum_^N {\left\lvert {{\left\langle {x},~{u_n} \right\rangle} } \right\rvert}^2 \ &= {\left\lVert {x} \right\rVert}^2 + \sum_{n, m \leq N}{\left\langle {x},~{u_n} \right\rangle} {\overline{{{\left\langle {x},~{u_m} \right\rangle} }}}{\left\langle {u_n},~{u_m} \right\rangle}
  • 2\sum_^N {\left\lvert {{\left\langle {x},~{u_n} \right\rangle} } \right\rvert}^2 \ &= {\left\lVert {x} \right\rVert}^2 + \sum_{n, m\leq N} {\left\langle {x},~{u_n} \right\rangle} {\overline{{{\left\langle {x},~{u_m} \right\rangle}}}} \delta_{mn}
  • 2\sum_^N {\left\lvert {{\left\langle {x},~{u_n} \right\rangle} } \right\rvert}^2 \ &= {\left\lVert {x} \right\rVert}^2 + \sum_{n\leq N} {\left\lvert {{\left\langle {x},~{u_n} \right\rangle}} \right\rvert}^2
  • 2\sum_^N {\left\lvert {{\left\langle {x},~{u_n} \right\rangle} } \right\rvert}^2 \ &= {\left\lVert {x} \right\rVert}^2
  • \sum_^N {\left\lvert {{\left\langle {x},~{u_n} \right\rangle} } \right\rvert}^2 .\end{align*} `{=html}

• Now take \(\lim_{N\to\infty}\) and use that \({\left\lVert {{-}} \right\rVert}\) is continuous.

• Set \begin{align*} x\coloneqq\sum_{n\in {\mathbb{N}}} a_n u_n .\end{align*}

• Checking the first desired property: \begin{align*} {\left\langle {x},~{u_m} \right\rangle} &= {\left\langle { \sum_{n\geq 1} a_n u_n },~{u_m} \right\rangle} \\ &=\sum_{n\geq 1} a_n {\left\langle { u_n },~{u_m} \right\rangle} \\ &=\sum_{n\geq 1} a_n \delta_{mn} \\ &= a_m .\end{align*}

• That \(x\in H\): this would follow from \begin{align*} {\left\lVert {x} \right\rVert}^2 = \sum_n {\left\lvert {{\left\langle {x},~{u_n } \right\rangle}} \right\rvert}^2 = \sum_n {\left\lvert {a_n} \right\rvert}^2 <\infty .\end{align*} The inequality holds by assumption since \(\left\{{a_n}\right\}\in\ell^2\), so it suffices to show the first equality:

\begin{align*} {\left\lVert {x} \right\rVert}^2 &\coloneqq{\left\langle {x},~{x} \right\rangle} \\ &= {\left\langle {\sum_n a_n u_n},~{\sum_m a_m u_m} \right\rangle} \\ &= \sum_{n, m} a_n {\overline{{a_m}}} {\left\langle {u_n},~{u_m} \right\rangle} \\ &= \sum_{n, m} a_n {\overline{{a_m}}} \delta_{mn} \\ &= \sum_{n} a_n {\overline{{a_n}}} \\ &= \sum_{n} {\left\lvert {a_n} \right\rvert}^2 \\ &= \sum_n {\left\lvert {{\left\langle {x},~{u_n} \right\rangle}} \right\rvert}^2 .\end{align*}

• Note \(X = [0, 1] \implies m(X) = 1\).

• By Holder’s inequality with \(p=q=2\), \begin{align*} {\left\lVert {f} \right\rVert}_1 = {\left\lVert {f\cdot 1} \right\rVert}_1 \leq {\left\lVert {f} \right\rVert}_2 \cdot {\left\lVert {1} \right\rVert}_2 = {\left\lVert {f} \right\rVert}_2 \cdot m(X)^{\frac 1 2} = {\left\lVert {f} \right\rVert}_2, \end{align*}

• Thus \(L^2(X) \subseteq L^1(X)\)

• Since they share a common dense subset (simple functions), \(L^2\) is dense in \(L^1\)

Let \(f\in L^2\subseteq L^1\) be arbitrary.

Claim: \(\Lambda\in L^1(X) {}^{ \vee }\implies \Lambda \in L^2(X) {}^{ \vee }\).

• Suffices to show that \({\left\lVert {\Gamma} \right\rVert}_{L^2(X) {}^{ \vee }} \coloneqq\sup_{{\left\lVert {f} \right\rVert}_2 = 1} {\left\lvert {\Gamma(f)} \right\rvert} < \infty\), since bounded implies continuous.

• By the lemma, \({\left\lVert {f} \right\rVert}_1 \leq C{\left\lVert {f} \right\rVert}_2\) for some constant \(C \approx m(X)\).

• Note \begin{align*}{\left\lVert {\Lambda} \right\rVert}_{L^1(X) {}^{ \vee }} \coloneqq\displaystyle\sup_{{\left\lVert {f} \right\rVert}_1 = 1} {\left\lvert {\Lambda(f)} \right\rvert}\end{align*}

• Define \(\widehat{f} = {f\over {\left\lVert {f} \right\rVert}_1}\) so \({\left\lVert {\widehat{f}} \right\rVert}_1 = 1\)

• Since \({\left\lVert {\Lambda} \right\rVert}_{1 {}^{ \vee }}\) is a supremum over all \(f \in L^1(X)\) with \({\left\lVert {f} \right\rVert}_1 =1\), \begin{align*} {\left\lvert {\Lambda(\widehat{f})} \right\rvert} \leq {\left\lVert {\Lambda} \right\rVert}_{(L^1(X)) {}^{ \vee }} ,\end{align*}

• Then \begin{align*} \frac{{\left\lvert {\Lambda(f)} \right\rvert}}{{\left\lVert {f} \right\rVert}_1} &= {\left\lvert {\Lambda(\widehat{f})} \right\rvert} \leq {\left\lVert {\Lambda} \right\rVert}_{L^1(X) {}^{ \vee }} \\ \implies {\left\lvert {\Lambda(f)} \right\rvert} &\leq {\left\lVert {\Lambda} \right\rVert}_{1 {}^{ \vee }} \cdot {\left\lVert {f} \right\rVert}_1 \\ &\leq {\left\lVert {\Lambda} \right\rVert}_{1 {}^{ \vee }} \cdot C {\left\lVert {f} \right\rVert}_2 < \infty \quad\text{by assumption} ,\end{align*}

• So \(\Lambda \in (L^2) {}^{ \vee }\).

Now apply Riesz Representation for \(L^2\): there is a \(g \in L^2\) such that \begin{align*}f\in L^2 \implies \Lambda(f) = {\left\langle {f},~{g} \right\rangle} \coloneqq\int_0^1 f(x) \overline{g(x)}\, dx.\end{align*}

• It suffices to show \({\left\lVert {g} \right\rVert}_{L^\infty(X)} < \infty\).

• Since we’re assuming \({\left\lVert {\Gamma} \right\rVert}_{L^1(X) {}^{ \vee }} < \infty\), it suffices to show the stated equality.

  \todo[inline]{Is this assumed..? Or did we show it..?}

• Claim: \({\left\lVert {\Lambda} \right\rVert}_{L^1(X) {}^{ \vee }} ={\left\lVert {g} \right\rVert}_{L^\infty(X)}\)

  • The result will follow since \(\Lambda\) was assumed to be in \(L^1(X) {}^{ \vee }\), so \({\left\lVert {\Lambda} \right\rVert}_{L^1(X) {}^{ \vee }} < \infty\).

  • \(\leq\): \begin{align*} {\left\lVert {\Lambda} \right\rVert}_{L^1(X) {}^{ \vee }} &= \sup_{{\left\lVert {f} \right\rVert}_1 = 1} {\left\lvert {\Lambda(f)} \right\rvert} \\ &= \sup_{{\left\lVert {f} \right\rVert}_1 = 1} {\left\lvert {\int_X f \overline{g}} \right\rvert} \quad\text{by (i)}\\ &= \sup_{{\left\lVert {f} \right\rVert}_1 = 1} \int_X {\left\lvert {f \overline{g}} \right\rvert} \\ &\coloneqq\sup_{{\left\lVert {f} \right\rVert}_1 = 1} {\left\lVert {fg} \right\rVert}_1 \\ &\leq \sup_{{\left\lVert {f} \right\rVert}_1 = 1} {\left\lVert {f} \right\rVert}_1 {\left\lVert {g} \right\rVert}_\infty \quad\text{by Holder with } p=1,q=\infty\\ &= {\left\lVert {g} \right\rVert}_\infty ,\end{align*}

  • \(\geq\):

    • Suppose toward a contradiction that \({\left\lVert {g} \right\rVert}_\infty > {\left\lVert {\Lambda} \right\rVert}_{1 {}^{ \vee }}\).

    • Then there exists some \(E\subseteq X\) with \(m(E) > 0\) such that \begin{align*}x\in E \implies {\left\lvert {g(x)} \right\rvert} > {\left\lVert {\Lambda} \right\rVert}_{L^1(X) {}^{ \vee }}.\end{align*}

    • Define \begin{align*} h = \frac{1}{m(E)} \frac{\overline{g}}{{\left\lvert {g} \right\rvert}} \chi_E .\end{align*}

    • Note \({\left\lVert {h} \right\rVert}_{L^1(X)} = 1\).

    • Then \begin{align*} \Lambda(h) &= \int_X hg \\ &\coloneqq\int_X \frac{1}{m(E)} \frac{g \overline g}{{\left\lvert {g} \right\rvert}} \chi_E \\ &= \frac{1}{m(E)} \int_E {\left\lvert {g} \right\rvert} \\ &\geq \frac{1}{m(E)} {\left\lVert {g} \right\rVert}_\infty m(E) \\ &= {\left\lVert {g} \right\rVert}_\infty \\ &> {\left\lVert {\Lambda} \right\rVert}_{L^1(X) {}^{ \vee }} ,\end{align*} a contradiction since \({\left\lVert {\Lambda} \right\rVert}_{L^1(X) {}^{ \vee }}\) is the supremum over all \(h_\alpha\) with \({\left\lVert {h_\alpha} \right\rVert}_{L^1(X)} = 1\).

• Let \(\left\{{f_n}\right\}\) be a Cauchy sequence in \(C(I, {\left\lVert {{-}} \right\rVert}_\infty)\), so \(\lim_n\lim_m {\left\lVert {f_m - f_n} \right\rVert}_\infty = 0\), we will show it converges to some \(f\) in this space.

• For each fixed \(x_0 \in [0, 1]\), the sequence of real numbers \(\left\{{f_n(x_0)}\right\}\) is Cauchy in \({\mathbf{R}}\) since \begin{align*} x_0\in I \implies {\left\lvert {f_m(x_0) - f_n(x_0)} \right\rvert} \leq \sup_{x\in I} {\left\lvert {f_m(x) - f_n(x)} \right\rvert} \coloneqq{\left\lVert {f_m - f_n} \right\rVert}_\infty \overset{m>n\to\infty}\to 0, \end{align*}

• Since \({\mathbf{R}}\) is complete, this sequence converges and we can define \(f(x) \coloneqq\lim_{k\to \infty} f_n(x)\).

• Thus \(f_n\to f\) pointwise by construction

• Claim: \({\left\lVert {f - f_n} \right\rVert} \overset{n\to\infty}\to 0\), so \(f_n\) converges to \(f\) in \(C([0, 1], {\left\lVert {{-}} \right\rVert}_\infty)\).

  • Proof:
    • Fix \({\varepsilon}> 0\); we will show there exists an \(N\) such that \(n\geq N \implies {\left\lVert {f_n - f} \right\rVert} < {\varepsilon}\)
    • Fix an \(x_0 \in I\). Since \(f_n \to f\) pointwise, choose \(N_1\) large enough so that \begin{align*}n\geq N_1 \implies {\left\lvert {f_n(x_0) - f(x_0)} \right\rvert} < {\varepsilon}/2.\end{align*}
    • Since \({\left\lVert {f_n - f_m} \right\rVert}_\infty \to 0\), choose and \(N_2\) large enough so that \begin{align*}n, m \geq N_2 \implies {\left\lVert {f_n - f_m} \right\rVert}_\infty < {\varepsilon}/2.\end{align*}
    • Then for \(n, m \geq \max(N_1, N_2)\), we have \begin{align*} {\left\lvert {f_n(x_0) - f(x_0)} \right\rvert} &= {\left\lvert {f_n(x_0) - f(x_0) + f_m(x_0) - f_m(x_0)} \right\rvert} \\ &= {\left\lvert {f_n(x_0) - f_m(x_0) + f_m(x_0) - f(x_0)} \right\rvert} \\ &\leq {\left\lvert {f_n(x_0) - f_m(x_0)} \right\rvert} + {\left\lvert {f_m(x_0) - f(x_0)} \right\rvert} \\ &< {\left\lvert {f_n(x_0) - f_m(x_0)} \right\rvert} + {{\varepsilon}\over 2} \\ &\leq \sup_{x\in I} {\left\lvert {f_n(x) - f_m(x)} \right\rvert} + {{\varepsilon}\over 2} \\ &< {\left\lVert {f_n - f_m} \right\rVert}_\infty + {{\varepsilon}\over 2} \\ &\leq {{\varepsilon}\over 2} + {{\varepsilon}\over 2} \\ \implies {\left\lvert {f_n(x_0) - f(x_0)} \right\rvert} &< {\varepsilon}\\ \implies \sup_{x\in I} {\left\lvert {f_n(x_0) - f(x_0)} \right\rvert} &\leq \sup_{x\in I} {\varepsilon}\quad\text{by order limit laws} \\ \implies {\left\lVert {f_n - f} \right\rVert} &\leq {\varepsilon}\\ .\end{align*}

• \(f\) is the uniform limit of continuous functions and thus continuous, so \(f\in C([0, 1])\).

• It suffices to produce a Cauchy sequence that does not converge to a continuous function.

• Take the following sequence of functions:

  • \(f_1\) increases linearly from 0 to 1 on \([0, 1/2]\) and is 1 on \([1/2, 1]\)
  • \(f_2\) is 0 on \([0, 1/4]\) increases linearly from 0 to 1 on \([1/4, 1/2]\) and is 1 on \([1/2, 1]\)
  • \(f_3\) is 0 on \([0, 3/8]\) increases linearly from 0 to 1 on \([3/8, 1/2]\) and is 1 on \([1/2, 1]\)
  • \(f_3\) is 0 on \([0, (1/2 - 3/8)/2]\) increases linearly from 0 to 1 on \([(1/2 - 3/8)/2, 1/2]\) and is 1 on \([1/2, 1]\)

  Idea: take sequence starting points for the triangles: \(0, 0 + {1\over 4}, 0 + {1 \over 4} + {1\over 8}, \cdots\) which converges to \(1/2\) since \(\sum_{k=1}^\infty{1\over 2^k} = -{1\over 2} + \sum_{k=0}^\infty {1\over 2^k}\).

• Then each \(f_n\) is clearly integrable, since its graph is contained in the unit square.

• \(\left\{{f_n}\right\}\) is Cauchy: geometrically subtracting areas yields a single triangle whose area tends to 0.

• But \(f_n\) converges to \(\chi_{[{1\over 2}, 1]}\) which is discontinuous.

\todo[inline]{show that $\int_0^1 {\left\lvert {f_n(x) - f_m(x)} \right\rvert} \,dx \to 0$ rigorously, show that no $g\in L^1([0, 1])$ can converge to this indicator function.}