Question

Consider the metric spaces \((C[0,1],d_1)\) and \((C[0,1],d_\infty)\), under the metrics \(d_1(f,g)=\int_0^1 |f(x)-g(x)|\,dx\) and \(d_\infty(f,g)=\sup_{x\in[0,1]}\{|f(x)-g(x)|\}\), for \(f,g\in C[0,1]\). Then which one of the options given below is correct?

• A. The metrics \(d_1\) and \(d_\infty\) are equivalent.
• B. Every open ball in \((C[0,1],d_1)\) is open in \((C[0,1],d_\infty)\).
• C. \((C[0,1],d_1)\) is a complete metric space.
• D. \((C[0,1],d_\infty)\) is a complete metric space.

Hint:

Compare d1 and d_infty using a tall thin spike function, and recall that uniform limits of continuous functions stay continuous.

Answer 1

The Correct Option is D

Step 1: First note \(d_1(f,g)=\int_0^1|f-g|\,dx\le \int_0^1 d_\infty(f,g)\,dx=d_\infty(f,g)\), so \(d_1\le d_\infty\) always.

Step 2 (Option A): Equivalence would require that a sequence converges in \(d_1\) exactly when it converges in \(d_\infty\). Take a spike \(f_n\) that is a triangle of height \(n\) and base \(1/n^2\) sitting near \(x=0\). Then \(d_\infty(f_n,0)=n\to\infty\), so \(f_n\) does not converge to \(0\) in \(d_\infty\), but \(d_1(f_n,0)=\tfrac12\cdot n\cdot \tfrac{1}{n^2}=\tfrac{1}{2n}\to 0\), so \(f_n\to 0\) in \(d_1\). The two metrics give different convergent sequences, so they are not equivalent; (A) is false.

Step 3 (Option C): Let \(f_n\) be \(0\) on \([0,\tfrac12]\), rise linearly to \(1\) on \([\tfrac12,\tfrac12+\tfrac1n]\), and stay at \(1\) on \([\tfrac12+\tfrac1n,1]\). Since \(d_1(f_n,f_m)\) is bounded by the area of a shrinking triangle, \((f_n)\) is Cauchy in \(d_1\). But \(f_n\to \chi\), the step function that is \(0\) on \([0,\tfrac12]\) and \(1\) on \((\tfrac12,1]\), in the \(d_1\) sense, and \(\chi\) is discontinuous, so it is not in \(C[0,1]\). No continuous function can be the \(d_1\)-limit, so \((C[0,1],d_1)\) is not complete; (C) is false.

Step 4 (Option D): If \((f_n)\) is Cauchy in \(d_\infty\), then for every \(x\), \((f_n(x))\) is a Cauchy sequence of real numbers, so it converges pointwise to some \(f(x)\); the uniform Cauchy condition upgrades this to uniform convergence \(f_n\to f\). A uniform limit of continuous functions on \([0,1]\) is continuous (the standard \(\varepsilon/3\) argument), so \(f\in C[0,1]\) and \(d_\infty(f_n,f)\to 0\). Hence \((C[0,1],d_\infty)\) is complete, matching the classical fact that \(C[0,1]\) with the sup norm is a Banach space; (D) is true.

Remark: Since \(d_1\le d_\infty\) pointwise, one can also show every \(d_1\)-open ball is \(d_\infty\)-open, but that is a one-directional relation between the two topologies, not a statement about completeness, so the key distinguishing fact the question is built around is captured by option (D).

\[\boxed{\text{Option (D): } (C[0,1],d_\infty) \text{ is complete}}\]