Question

Let \( X \) be the space of all continuously differentiable real valued functions on \( [0,1] \). Define the following norms on \( X \):
\[ p_1(x) = \sup\{ |x(t)| : t \in [0,1] \} \]
\[ p_2(x) = \sup\left\{ \left| \frac{d}{dt} x(t) \right| : t \in [0,1] \right\} \]
and \( p_3(x) = p_1(x) + p_2(x) \).
Which one of the following is TRUE?

• A. \( (X, p_1) \) is a Banach space.
• B. \( (X, p_2) \) is a Banach space.
• C. \( (X, p_3) \) is not a Banach space.
• D. \( (X, p_3) \) is a Banach space.

Hint:

Check whether each candidate is complete, and first check whether p2 alone is even a valid norm on constant functions.

Answer 1

The Correct Option is D

Step 1: Understand the space and the three norms.
\( X \) is the set of all continuously differentiable real functions on \( [0,1] \), often written \( C^1[0,1] \). The three quantities given are
\[ p_1(x) = \sup_{t \in [0,1]} |x(t)|, \qquad p_2(x) = \sup_{t \in [0,1]} |x'(t)|, \qquad p_3(x) = p_1(x) + p_2(x) \]
We need to decide which of the four statements about completeness, meaning being a Banach space, a complete normed space, is true.

Step 2: Test p1 for completeness.
\( p_1 \) is just the usual sup norm, applied only to functions that happen to be \( C^1 \). Take a sequence of smooth functions \( x_n \) that converges uniformly to a function like \( g(t) = |t - \tfrac12| \), which is continuous but has a sharp corner at \( t = \tfrac12 \) and so is not differentiable there.
This sequence is Cauchy in \( p_1 \), since it converges uniformly, but its limit \( g \) does not belong to \( X \), because \( g \) is not continuously differentiable. So this Cauchy sequence has no limit inside \( X \) under \( p_1 \). This means \( (X, p_1) \) is not complete, so option (A) is false.

Step 3: Test p2, and notice it is not even a genuine norm.
A norm must be zero only for the zero function. Check \( p_2 \) on a nonzero constant function, say \( x(t) = 5 \) for all \( t \). Its derivative is \( x'(t) = 0 \), so
\[ p_2(x) = \sup_{t}|x'(t)| = 0 \]
even though \( x \) is not the zero function. So \( p_2 \) fails the positive-definiteness property of a norm; it is only a seminorm. Since \( (X, p_2) \) is not even a genuine normed space, it cannot be called a Banach space in the standard sense. This rules out option (B).

Step 4: Test p3, the combined norm.
\( p_3(x) = p_1(x) + p_2(x) \) is the standard \( C^1 \) norm: it is zero only when both \( \sup|x| = 0 \) and \( \sup|x'| = 0 \), which forces \( x \equiv 0 \). So unlike \( p_2 \) alone, \( p_3 \) is a genuine norm.
Now take any sequence \( x_n \) that is Cauchy in \( p_3 \). This means both \( x_n \) and \( x_n' \) are uniformly Cauchy, so both converge uniformly: \( x_n \to g \) and \( x_n' \to h \) for some continuous \( g, h \).
A standard real-analysis fact says that if \( x_n \to g \) uniformly and \( x_n' \to h \) uniformly, then \( g \) is differentiable with \( g' = h \). So \( g \) is continuously differentiable, meaning \( g \in X \), and \( x_n \to g \) in the \( p_3 \) norm since both parts converge uniformly.
So every \( p_3 \)-Cauchy sequence converges to a limit inside \( X \). This means \( (X, p_3) \) is complete, so it is a Banach space. This makes option (D) true and rules out option (C).

Final Answer:
Only the combined norm \( p_3 = p_1 + p_2 \), which controls both the function and its derivative uniformly, makes \( X = C^1[0,1] \) complete. \[ \boxed{\text{(D) } (X, p_3) \text{ is a Banach space.}} \]