Let \(C([0,1])\) denote the space of continuous functions on \([0,1]\), endowed with the supremum norm. Show that \(C([0,1])\) is a Banach space.
Let \(B_p=L^p([0,1])\), with \(1<p<\infty\). Define weak and strong convergence in \(B_p\). Then show that \[ f_n(x)=\sin(nx) \] converges weakly to \(0\), but not strongly to \(0\), in \(B_2\).
Solution
1. \(C([0,1])\) is complete in the supremum norm.
Let \((u_n)\) be Cauchy in \(\|\cdot\|_\infty\). For each \(x\in[0,1]\), the scalar sequence \((u_n(x))\) is Cauchy because \[ |u_n(x)-u_m(x)|\le \|u_n-u_m\|_\infty. \] Define \[ u(x)=\lim_{n\to\infty}u_n(x). \] Given \(\varepsilon>0\), choose \(N\) such that \[ \|u_n-u_m\|_\infty<\varepsilon \qquad(n,m\ge N). \] Fix \(n\ge N\) and let \(m\to\infty\). Then for every \(x\), \[ |u_n(x)-u(x)|\le\varepsilon, \] so \[ \|u_n-u\|_\infty\le\varepsilon. \] Thus \(u_n\to u\) uniformly. A uniform limit of continuous functions is continuous, hence \(u\in C([0,1])\). Therefore \(C([0,1])\) is complete.
2. Strong and weak convergence in \(L^p\).
Strong convergence \(v_n\to v\) in \(L^p([0,1])\) means \[ \|v_n-v\|_p\longrightarrow0. \] If \(1<p<\infty\) and \(q\) is the conjugate exponent, \(1/p+1/q=1\), weak convergence \(v_n\rightharpoonup v\) means \[ \int_0^1 v_n(x)g(x)\,dx \longrightarrow \int_0^1 v(x)g(x)\,dx \qquad(g\in L^q([0,1])). \] For \(p=2\), this is convergence against every element of \(L^2([0,1])\) under the Hilbert-space inner product.
3. \(\sin(nx)\rightharpoonup0\) in \(L^2([0,1])\).
Let \(g\in L^2([0,1])\). Since \([0,1]\) has finite measure, Hölder’s inequality gives \(g\in L^1([0,1])\). By the Riemann–Lebesgue lemma, \[ \int_0^1 g(x)e^{inx}\,dx\longrightarrow0. \] Taking imaginary parts, \[ \int_0^1 g(x)\sin(nx)\,dx\longrightarrow0. \] Hence \(f_n\rightharpoonup0\) in \(L^2([0,1])\).
4. The convergence is not strong.
We have \[ \|f_n\|_2^2 =\int_0^1\sin^2(nx)\,dx =\frac12-\frac{\sin(2n)}{4n}. \] Therefore \[ \|f_n\|_2^2\longrightarrow\frac12, \] so \(\|f_n\|_2\) does not tend to \(0\). Thus \(f_n\) does not converge strongly to \(0\) in \(L^2([0,1])\).