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?

Question

Let $A$ be a square matrix. If $A^2 = A$, then the matrix $A$ is called:

Answer 1

This question is really about testing completeness of $C^1[0,1]$ under three different ways of measuring size. Let's go option by option with concrete examples first, then confirm with a general argument for $p_3$.

(A) (X, p1) is a Banach space: Pick a sequence of smooth functions $x_n(t)$ that converges uniformly to the tent function $g(t) = |t - 1/2|$, for example by smoothing the corner more and more sharply as $n$ grows. Each $x_n$ is in $X$ and the sequence is Cauchy under the sup norm $p_1$, since it converges uniformly. But $g$ is not differentiable at $t = 1/2$, so $g \notin X$. A Cauchy sequence with no limit inside $X$ means $(X, p_1)$ is not complete. This option is false.
(B) (X, p2) is a Banach space: Before even asking about completeness, check whether $p_2$ qualifies as a norm at all. Take the nonzero constant function $x(t) = 1$. Its derivative is $0$ everywhere, so $p_2(x) = \sup|x'(t)| = 0$, even though $x \neq 0$. A norm must be zero only for the zero vector, so $p_2$ fails this basic requirement and is only a seminorm. Since $(X, p_2)$ is not a genuine normed space to begin with, calling it a Banach space is not correct. This option is false.
(C) (X, p3) is not a Banach space: The next point shows $(X, p_3)$ is in fact complete, so this option is false.
(D) (X, p3) is a Banach space: Let $(x_n)$ be Cauchy in $p_3 = p_1 + p_2$. Since $p_1(x_n - x_m) \to 0$ and $p_2(x_n - x_m) \to 0$ as $n,m \to \infty$, both $x_n$ and its derivative $x_n'$ are uniformly Cauchy sequences of continuous functions on $[0,1]$. By completeness of the space of continuous functions with the sup norm, $x_n \to g$ uniformly and $x_n' \to h$ uniformly for some continuous $g$ and $h$. A standard theorem on uniform convergence of derivatives then tells us $g$ is differentiable and $g' = h$, so $g \in C^1[0,1] = X$. Since both $x_n \to g$ and $x_n' \to g'$ uniformly, $p_3(x_n - g) \to 0$, so $x_n \to g$ in the $p_3$ norm. Every Cauchy sequence converges inside $X$, so $(X, p_3)$ is complete, that is, a Banach space. This option is true.

Let's summarize:

$p_1$ alone is not enough because the uniform limit of $C^1$ functions need not be differentiable.
$p_2$ alone is not even a valid norm, since it vanishes on every nonzero constant function.
Only $p_3 = p_1 + p_2$, which tracks the function and its derivative together, is both a genuine norm and complete.

\[ \boxed{\text{(D) } (X, p_3) \text{ is a Banach space}} \]

Answer 2

This question is really about testing completeness of $C^1[0,1]$ under three different ways of measuring size. Let's go option by option with concrete examples first, then confirm with a general argument for $p_3$.

(A) (X, p1) is a Banach space: Pick a sequence of smooth functions $x_n(t)$ that converges uniformly to the tent function $g(t) = |t - 1/2|$, for example by smoothing the corner more and more sharply as $n$ grows. Each $x_n$ is in $X$ and the sequence is Cauchy under the sup norm $p_1$, since it converges uniformly. But $g$ is not differentiable at $t = 1/2$, so $g \notin X$. A Cauchy sequence with no limit inside $X$ means $(X, p_1)$ is not complete. This option is false.
(B) (X, p2) is a Banach space: Before even asking about completeness, check whether $p_2$ qualifies as a norm at all. Take the nonzero constant function $x(t) = 1$. Its derivative is $0$ everywhere, so $p_2(x) = \sup|x'(t)| = 0$, even though $x \neq 0$. A norm must be zero only for the zero vector, so $p_2$ fails this basic requirement and is only a seminorm. Since $(X, p_2)$ is not a genuine normed space to begin with, calling it a Banach space is not correct. This option is false.
(C) (X, p3) is not a Banach space: The next point shows $(X, p_3)$ is in fact complete, so this option is false.
(D) (X, p3) is a Banach space: Let $(x_n)$ be Cauchy in $p_3 = p_1 + p_2$. Since $p_1(x_n - x_m) \to 0$ and $p_2(x_n - x_m) \to 0$ as $n,m \to \infty$, both $x_n$ and its derivative $x_n'$ are uniformly Cauchy sequences of continuous functions on $[0,1]$. By completeness of the space of continuous functions with the sup norm, $x_n \to g$ uniformly and $x_n' \to h$ uniformly for some continuous $g$ and $h$. A standard theorem on uniform convergence of derivatives then tells us $g$ is differentiable and $g' = h$, so $g \in C^1[0,1] = X$. Since both $x_n \to g$ and $x_n' \to g'$ uniformly, $p_3(x_n - g) \to 0$, so $x_n \to g$ in the $p_3$ norm. Every Cauchy sequence converges inside $X$, so $(X, p_3)$ is complete, that is, a Banach space. This option is true.

Let's summarize:

$p_1$ alone is not enough because the uniform limit of $C^1$ functions need not be differentiable.
$p_2$ alone is not even a valid norm, since it vanishes on every nonzero constant function.
Only $p_3 = p_1 + p_2$, which tracks the function and its derivative together, is both a genuine norm and complete.

\[ \boxed{\text{(D) } (X, p_3) \text{ is a Banach space}} \]

Show Hint

The Correct Option is D

Solution and Explanation

Questions Asked in GATE MA exam