Normed and Banach Spaces Questions and Answers

Q: Consider the statementsP: If \(X\) is a normed linear space and \(M \subseteq X\) is a subspace, then the closure \(\bar{M}\) is also a subspace of \(X\).Q: If \(X\) is a Banach space and \(\sum x_{n}\) is an absolutely convergent series in \(X\), then \(\sum x_{n}\) is convergent.R: Let \(M_{1}\) and \(M_{2}\) be subspaces of an inner product space such that \(M_{1} \cap M_{2} = \{0\}\). Then \(\forall m_{1} \in M_{1},\ m_{2} \in M_{2}\); \(\|m_{1} + m_{2}\|^{2} = \|m_{1}\|^{2} + \|m_{2}\|^{2}\).S: Let \(f : X \to Y\) be a linear transformation from the Banach Space \(X\) into the Banach space \(Y\). If \(f\) is continuous, then the graph of \(f\) is always compact.The correct statements amongst the above are:

The correct statements are P and Q only.P is true: if \(x_{n} \to x\) and \(y_{n} \to y\) with \(x_{n}, y_{n} \in M\), then \(\alpha x_{n} + \beta y_{n} \to \alpha x + \beta y\), so the closure of a subspace is a subspace. Q is true and in fact characterises Banach spaces: the partial sums of an absolutely convergent series form a Cauchy sequence, which converges by completeness.R is false: the Pythagorean identity needs \(M_{1} \perp M_{2}\), not merely \(M_{1} \cap M_{2} = \{0\}\). In \(\mathbb{R}^{2}\), with \(m_{1} = (1, 0)\) and \(m_{2} = (1, 1)\) on two different lines, \(\|m_{1} + m_{2}\|^{2} = 5\) but \(\|m_{1}\|^{2} + \|m_{2}\|^{2} = 3\). S is false: the graph of a continuous linear map is closed, but it contains \((tx, tf(x))\) for every scalar \(t\), so it is unbounded and never compact unless \(X = \{0\}\).

Q: Let \(V = C^{1}[0,1]\), \(X = (C[0,1], \|\ \|_{\infty})\) and \(Y = (C[0,1], \|\ \|_{2})\). Then \(V\) is

\(V\) is dense in both \(X\) and \(Y\).By the Weierstrass approximation theorem, polynomials are dense in \(C[0,1]\) under the supremum norm. Polynomials are continuously differentiable, so \(C^{1}[0,1]\), which contains them, is dense in \(X\).For \(Y\), note that \(\|f\|_{2} = \left(\int_{0}^{1} |f|^{2}\right)^{1/2} \le \|f\|_{\infty}\). So any approximation that is close in the supremum norm is at least as close in the \(L^{2}\) norm, and \(V\) is dense in \(Y\) as well.That rules out every option that denies density in either space.

Q: Let X be a real normed linear space. Let \(X_{0} = \{x \in X : \|x\| = 1\}\). If \(X_{0}\) contains two distinct points x and y and the line segment joining them, then, which of the following statements is TRUE?

The true statement is \(\|x + y\| = \|x\| + \|y\|\) and x, y are linearly independent.The midpoint (x + y)/2 lies on the segment, so it is in \(X_{0}\) and has norm 1. Hence \(\|x + y\| = 2 = \|x\| + \|y\|\).If x and y were dependent, then y = cx with |c| = 1 since both have norm 1, and c = -1 because they are distinct. The midpoint would then be 0, of norm 0, not 1. So x and y are independent.The third option fails because \(\|x+y\|^{2} = 4\) while \(\|x\|^{2} + \|y\|^{2} = 2\). The second and fourth claim linear dependence, which was just ruled out.

Q: Let C[0,1] denote the set of all real valued continuous functions defined on [0,1] and \(\|f\|_{\infty} = \sup\{|f(x)| : x \in [0,1]\}\) for all \(f \in C[0,1]\). Let\[ X = \{f \in C[0,1] : f(0) = f(1) = 0\}. \]Define \(F : (C[0,1], \|\cdot\|_{\infty}) \to \mathbb{R}\) by \(F(f) = \int_{0}^{1}f(t)\,dt\) for all \(f \in C[0,1]\).Denote \(S_{X} = \{f \in X : \|f\|_{\infty} = 1\}\).Then the set \(\{f \in X : F(f) = \|F\|\} \cap S_{X}\) has

The set has NO element.F is integration over [0,1], and \(|F(f)| \le \|f\|_{\infty}\), with equality for the constant function 1, so \(\|F\| = 1\). Functions in X can come arbitrarily close to this value: a function equal to 1 except on short ramps near 0 and 1 has integral close to 1.But no \(f \in S_{X}\) attains it. If \(\|f\|_{\infty} = 1\) and \(\int_{0}^{1}f = 1\), then \(\int_{0}^{1}(1-f) = 0\) with \(1 - f \ge 0\) continuous, forcing f = 1 everywhere, which contradicts f(0) = 0. So the set is empty.This is the standard example of a functional on the closed subspace X that does not attain its norm.

Q: Let\[ \begin{aligned} c_{00} = \{&(x_{1},x_{2},x_{3},\ldots) : x_{i} \in \mathbb{R},\ i \in \mathbb{N}, \\ &x_{i} \ne 0 \text{ only for finitely many indices } i\}. \end{aligned} \]For \((x_{1},x_{2},x_{3},\ldots) \in c_{00}\), let \(\|(x_{1},x_{2},x_{3},\ldots)\|_{\infty} = \sup\{|x_{i}| : i \in \mathbb{N}\}\).Define \(F, G : (c_{00}, \|\cdot\|_{\infty}) \to (c_{00}, \|\cdot\|_{\infty})\) by\[ \begin{aligned} F((x_{1},\ldots,x_{n},\ldots)) = (&(1+1)x_{1}, (2+\tfrac{1}{2})x_{2}, \\ &\ldots, (n+\tfrac{1}{n})x_{n}, \ldots), \end{aligned} \]\[ \begin{aligned} G((x_{1},\ldots,x_{n},\ldots)) = (&\tfrac{x_{1}}{1+1}, \tfrac{x_{2}}{2+\frac{1}{2}}, \\ &\ldots, \tfrac{x_{n}}{n+\frac{1}{n}}, \ldots), \end{aligned} \]for all \((x_{1},x_{2},\ldots,x_{n},\ldots) \in c_{00}\).Then

F is NOT continuous but G is continuous.A linear map between normed spaces is continuous exactly when it is bounded.F multiplies the n-th coordinate by n + 1/n. For the unit vector \(e_{n}\), \(\|F(e_{n})\|_{\infty} = n + 1/n\), which is unbounded while \(\|e_{n}\|_{\infty} = 1\). So F is unbounded and not continuous.G divides the n-th coordinate by n + 1/n, which is at least 2, so \(\|G(x)\|_{\infty} \le \tfrac{1}{2}\|x\|_{\infty}\). G is bounded and therefore continuous. (G is the inverse of F on \(c_{00}\), a bounded map whose inverse is unbounded.)

Q: Let\[ \begin{aligned} \ell^{p} = \Big\{ & x = (x_{n})_{n \ge 1} : x_{n} \in \mathbb{R}, \\ & \|x\|_{p} = \Big(\sum_{n=1}^{\infty}|x_{n}|^{p}\Big)^{1/p} \lt \infty \Big\} \end{aligned} \]for \(p = 1, 2\). Let\(C_{00} = \{(x_{n})_{n \ge 1} : x_{n} = 0\) for all but finitely many \(n \ge 1\}\).For \(x = (x_{n})_{n \ge 1} \in C_{00}\), define \(f(x) = \sum_{n=1}^{\infty}\frac{x_{n}}{\sqrt{n}}\). Consider the following statements.I. There exists a continuous linear functional \(F\) on \((\ell^{1}, \|\cdot\|_{1})\) such that \(F = f\) on \(C_{00}\).II. There exists a continuous linear functional \(G\) on \((\ell^{2}, \|\cdot\|_{2})\) such that \(G = f\) on \(C_{00}\).Which one of the following is correct?

I is TRUE and II is FALSE.I: for \(x \in C_{00}\), \(|f(x)| \le \sum|x_{n}|/\sqrt{n} \le \|x\|_{1}\), so \(f\) is bounded in the \(\ell^{1}\) norm. Since \(C_{00}\) is dense in \(\ell^{1}\), \(f\) extends to a continuous linear functional \(F\) on \(\ell^{1}\) (given by the bounded sequence \(1/\sqrt{n}\)).II: a continuous extension to \(\ell^{2}\) would be \(G(x) = \sum x_{n}c_{n}\) with \((c_{n}) \in \ell^{2}\), forcing \(c_{n} = 1/\sqrt{n}\). But \(\sum 1/n\) diverges, so \((1/\sqrt{n}) \notin \ell^{2}\). Concretely, \(x = (1/\sqrt{n})_{n \le N}\) has \(\|x\|_{2}^{2} = f(x) = \sum_{n \le N}1/n\), so \(f(x)/\|x\|_{2}\) is unbounded and \(f\) is not continuous on \(\ell^{2}\).

Q: Let \(X = \{f \in C[0,1] : f(0) = 0 = f(1)\}\) with the norm \(\|f\|_{\infty} = \sup_{0 \le t \le 1} |f(t)|\), where \(C[0,1]\) is the space of all real-valued continuous functions on [0, 1]. Let \(Y = C[0,1]\) with the norm \(\|f\|_{2} = \left(\int_{0}^{1} |f(t)|^{2} dt\right)^{1/2}\). Let \(U_{X}\) and \(U_{Y}\) be the closed unit balls in X and Y centred at the origin, respectively. Consider \(T: X \to \mathbb{R}\) and \(S: Y \to \mathbb{R}\) given by\[ Tf = \int_{0}^{1} f(t)\,dt \]and\[ Sf = \int_{0}^{1} f(t)\,dt. \]Consider the following statements:S1: \(\sup_{f \in U_{X}} |Tf|\) is attained at a point of \(U_{X}\).S2: \(\sup_{f \in U_{Y}} |Sf|\) is attained at a point of \(U_{Y}\).Then, which one of the following is correct?

S2 is TRUE and S1 is FALSE.S1: for f in \(U_{X}\), \(|Tf| \le \int_{0}^{1}|f| \le 1\), and functions equal to 1 except near the end points make \(Tf\) as close to 1 as we like, so the supremum is 1. But reaching 1 would need \(|f(t)| = 1\) everywhere, impossible for a continuous f with \(f(0) = f(1) = 0\). The supremum is not attained.S2: by the Cauchy-Schwarz inequality, \(|Sf| = |\int_{0}^{1} f \cdot 1| \le \|f\|_{2}\|1\|_{2} \le 1\), with equality for the constant function f = 1, which lies in \(U_{Y}\) since its norm is 1. The supremum 1 is attained. So only S2 holds.

Q: Consider the following two spaces:\(X = (C[-1,1], \|\cdot\|_{\infty})\), the space of all real-valued continuous functions defined on [-1, 1] equipped with the norm \(\|f\|_{\infty} = \sup_{t \in [-1,1]} |f(t)|\).\(Y = (C[-1,1], \|\cdot\|_{2})\), the space of all real-valued continuous functions defined on [-1, 1] equipped with the norm \(\|f\|_{2} = \left(\int_{-1}^{1} |f(t)|^{2} dt\right)^{1/2}\).Let W be the linear span over \(\mathbb{R}\) of all the Legendre polynomials. Then, which one of the following is correct?

W is dense in both X and Y.The Legendre polynomials \(P_{0}, P_{1}, P_{2}, \ldots\) have degrees 0, 1, 2, ..., so their span W is the set of all polynomials. By the Weierstrass approximation theorem, polynomials are dense in \(C[-1, 1]\) under the sup norm, so W is dense in X.For Y, note that \(\|f\|_{2} \le \sqrt{2}\|f\|_{\infty}\) on [-1, 1]. Any polynomial within \(\varepsilon\) of f in the sup norm is within \(\sqrt{2}\varepsilon\) in the 2-norm, so density in X carries over to Y. The weaker norm can only make it easier to approximate, never harder.

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

The true statement is \((X, p_3)\) is a Banach space.\(p_3\) is the usual \(C^1\) norm. If \((x_n)\) is Cauchy in \(p_3\), both \(x_n\) and \(x_n'\) converge uniformly, to some x and g; uniform convergence of the derivatives lets you pass to the limit in \(x_n(t) = x_n(0) + \int_0^t x_n'\), so x is continuously differentiable with \(x' = g\), and \(x_n \to x\) in \(p_3\). The space is complete.\(p_1\) alone fails: smooth functions can converge uniformly to \(|t - \frac{1}{2}|\), which is not differentiable, so \((X, p_1)\) is not complete. \(p_2\) vanishes on every constant function, so it is not even a norm and \((X, p_2)\) is not a Banach space; the option saying \(p_3\) fails is the negation of the true statement.

Q: Let\[ \begin{aligned} \ell^{\infty} = \{ & x = (x_n)_{n \geq 1} \mid x_n \in \mathbb{R}, \\ & \sup\{|x_n| : n = 1, 2, \ldots\} \lt \infty\} \end{aligned} \]with the supremum norm. Let \(T : \ell^{\infty} \to \ell^{\infty}\) be given by \(T(x_1, x_2, x_3, \ldots) = (x_1, \frac{x_2}{2}, \frac{x_3}{3}, \ldots)\).Which one of the following is TRUE?

The true statement is T is bounded and the inverse (from the range of T) exists but not bounded.T is bounded: \(|x_n / n| \le |x_n|\), so \(\|Tx\|_{\infty} \le \|x\|_{\infty}\) and \(\|T\| = 1\). It is one-to-one: \(Tx = 0\) forces every \(x_n = 0\). So the inverse exists on the range, and it is \((y_n) \mapsto (n y_n)\).That inverse is unbounded. The unit vector \(e_n\) is mapped by T to \(e_n / n\), of norm \(1/n\), and the inverse sends this back to \(e_n\), of norm 1, so\[ \frac{\|T^{-1}(e_n/n)\|}{\|e_n/n\|} = n \to \infty. \]So T is not 'not one-to-one', is not unbounded, and does not have a bounded inverse.

Consider the statements

P: If \(X\) is a normed linear space and \(M \subseteq X\) is a subspace, then the closure \(\bar{M}\) is also a subspace of \(X\).
Q: If \(X\) is a Banach space and \(\sum x_{n}\) is an absolutely convergent series in \(X\), then \(\sum x_{n}\) is convergent.
R: Let \(M_{1}\) and \(M_{2}\) be subspaces of an inner product space such that \(M_{1} \cap M_{2} = \{0\}\). Then \(\forall m_{1} \in M_{1},\ m_{2} \in M_{2}\); \(\|m_{1} + m_{2}\|^{2} = \|m_{1}\|^{2} + \|m_{2}\|^{2}\).
S: Let \(f : X \to Y\) be a linear transformation from the Banach Space \(X\) into the Banach space \(Y\). If \(f\) is continuous, then the graph of \(f\) is always compact.

The correct statements amongst the above are:

A.P and R only
B.Q and R only
C.P and Q only
D.R and S only

Exam Year(s)

GATE Mathematics - 2012,

Answer: C

The correct statements are P and Q only.

P is true: if \(x_{n} \to x\) and \(y_{n} \to y\) with \(x_{n}, y_{n} \in M\), then \(\alpha x_{n} + \beta y_{n} \to \alpha x + \beta y\), so the closure of a subspace is a subspace. Q is true and in fact characterises Banach spaces: the partial sums of an absolutely convergent series form a Cauchy sequence, which converges by completeness.

R is false: the Pythagorean identity needs \(M_{1} \perp M_{2}\), not merely \(M_{1} \cap M_{2} = \{0\}\). In \(\mathbb{R}^{2}\), with \(m_{1} = (1, 0)\) and \(m_{2} = (1, 1)\) on two different lines, \(\|m_{1} + m_{2}\|^{2} = 5\) but \(\|m_{1}\|^{2} + \|m_{2}\|^{2} = 3\). S is false: the graph of a continuous linear map is closed, but it contains \((tx, tf(x))\) for every scalar \(t\), so it is unbounded and never compact unless \(X = \{0\}\).

Discuss in Forum

Let \(V = C^{1}[0,1]\), \(X = (C[0,1], \|\ \|_{\infty})\) and \(Y = (C[0,1], \|\ \|_{2})\). Then \(V\) is

A.dense in \(X\) but NOT in \(Y\)
B.dense in \(Y\) but NOT in \(X\)
C.dense in both \(X\) and \(Y\)
D.neither dense in \(X\) nor dense in \(Y\)

Exam Year(s)

GATE Mathematics - 2015,

Answer: C

\(V\) is dense in both \(X\) and \(Y\).

By the Weierstrass approximation theorem, polynomials are dense in \(C[0,1]\) under the supremum norm. Polynomials are continuously differentiable, so \(C^{1}[0,1]\), which contains them, is dense in \(X\).

For \(Y\), note that \(\|f\|_{2} = \left(\int_{0}^{1} |f|^{2}\right)^{1/2} \le \|f\|_{\infty}\). So any approximation that is close in the supremum norm is at least as close in the \(L^{2}\) norm, and \(V\) is dense in \(Y\) as well.

That rules out every option that denies density in either space.

Discuss in Forum

Let X be a real normed linear space. Let \(X_{0} = \{x \in X : \|x\| = 1\}\). If \(X_{0}\) contains two distinct points x and y and the line segment joining them, then, which of the following statements is TRUE?

A.\(\|x + y\| = \|x\| + \|y\|\) and x, y are linearly independent
B.\(\|x + y\| = \|x\| + \|y\|\) and x, y are linearly dependent
C.\(\|x + y\|^{2} = \|x\|^{2} + \|y\|^{2}\) and x, y are linearly independent
D.\(\|x + y\| = 2\|x\|\|y\|\) and x, y are linearly dependent

Exam Year(s)

GATE Mathematics - 2022,

Answer: A

The true statement is \(\|x + y\| = \|x\| + \|y\|\) and x, y are linearly independent.

The midpoint (x + y)/2 lies on the segment, so it is in \(X_{0}\) and has norm 1. Hence \(\|x + y\| = 2 = \|x\| + \|y\|\).

If x and y were dependent, then y = cx with |c| = 1 since both have norm 1, and c = -1 because they are distinct. The midpoint would then be 0, of norm 0, not 1. So x and y are independent.

The third option fails because \(\|x+y\|^{2} = 4\) while \(\|x\|^{2} + \|y\|^{2} = 2\). The second and fourth claim linear dependence, which was just ruled out.

Discuss in Forum

Let C[0,1] denote the set of all real valued continuous functions defined on [0,1] and \(\|f\|_{\infty} = \sup\{|f(x)| : x \in [0,1]\}\) for all \(f \in C[0,1]\). Let

\[ X = \{f \in C[0,1] : f(0) = f(1) = 0\}. \]

Define \(F : (C[0,1], \|\cdot\|_{\infty}) \to \mathbb{R}\) by \(F(f) = \int_{0}^{1}f(t)\,dt\) for all \(f \in C[0,1]\).

Denote \(S_{X} = \{f \in X : \|f\|_{\infty} = 1\}\).

Then the set \(\{f \in X : F(f) = \|F\|\} \cap S_{X}\) has

A.NO element
B.exactly one element
C.exactly two elements
D.an infinite number of elements

Exam Year(s)

GATE Mathematics - 2023,

Answer: A

The set has NO element.

F is integration over [0,1], and \(|F(f)| \le \|f\|_{\infty}\), with equality for the constant function 1, so \(\|F\| = 1\). Functions in X can come arbitrarily close to this value: a function equal to 1 except on short ramps near 0 and 1 has integral close to 1.

But no \(f \in S_{X}\) attains it. If \(\|f\|_{\infty} = 1\) and \(\int_{0}^{1}f = 1\), then \(\int_{0}^{1}(1-f) = 0\) with \(1 - f \ge 0\) continuous, forcing f = 1 everywhere, which contradicts f(0) = 0. So the set is empty.

This is the standard example of a functional on the closed subspace X that does not attain its norm.

Discuss in Forum

Let

\[ \begin{aligned} c_{00} = \{&(x_{1},x_{2},x_{3},\ldots) : x_{i} \in \mathbb{R},\ i \in \mathbb{N}, \\ &x_{i} \ne 0 \text{ only for finitely many indices } i\}. \end{aligned} \]

For \((x_{1},x_{2},x_{3},\ldots) \in c_{00}\), let \(\|(x_{1},x_{2},x_{3},\ldots)\|_{\infty} = \sup\{|x_{i}| : i \in \mathbb{N}\}\).

Define \(F, G : (c_{00}, \|\cdot\|_{\infty}) \to (c_{00}, \|\cdot\|_{\infty})\) by

\[ \begin{aligned} F((x_{1},\ldots,x_{n},\ldots)) = (&(1+1)x_{1}, (2+\tfrac{1}{2})x_{2}, \\ &\ldots, (n+\tfrac{1}{n})x_{n}, \ldots), \end{aligned} \]

\[ \begin{aligned} G((x_{1},\ldots,x_{n},\ldots)) = (&\tfrac{x_{1}}{1+1}, \tfrac{x_{2}}{2+\frac{1}{2}}, \\ &\ldots, \tfrac{x_{n}}{n+\frac{1}{n}}, \ldots), \end{aligned} \]

for all \((x_{1},x_{2},\ldots,x_{n},\ldots) \in c_{00}\).

Then

A.F is continuous but G is NOT continuous
B.F is NOT continuous but G is continuous
C.both F and G are continuous
D.NEITHER F NOR G is continuous

Exam Year(s)

GATE Mathematics - 2023,

Answer: B

F is NOT continuous but G is continuous.

A linear map between normed spaces is continuous exactly when it is bounded.

F multiplies the n-th coordinate by n + 1/n. For the unit vector \(e_{n}\), \(\|F(e_{n})\|_{\infty} = n + 1/n\), which is unbounded while \(\|e_{n}\|_{\infty} = 1\). So F is unbounded and not continuous.

G divides the n-th coordinate by n + 1/n, which is at least 2, so \(\|G(x)\|_{\infty} \le \tfrac{1}{2}\|x\|_{\infty}\). G is bounded and therefore continuous. (G is the inverse of F on \(c_{00}\), a bounded map whose inverse is unbounded.)

Discuss in Forum

Let

\[ \begin{aligned} \ell^{p} = \Big\{ & x = (x_{n})_{n \ge 1} : x_{n} \in \mathbb{R}, \\ & \|x\|_{p} = \Big(\sum_{n=1}^{\infty}|x_{n}|^{p}\Big)^{1/p} \lt \infty \Big\} \end{aligned} \]

for \(p = 1, 2\). Let

\(C_{00} = \{(x_{n})_{n \ge 1} : x_{n} = 0\) for all but finitely many \(n \ge 1\}\).

For \(x = (x_{n})_{n \ge 1} \in C_{00}\), define \(f(x) = \sum_{n=1}^{\infty}\frac{x_{n}}{\sqrt{n}}\). Consider the following statements.

I. There exists a continuous linear functional \(F\) on \((\ell^{1}, \|\cdot\|_{1})\) such that \(F = f\) on \(C_{00}\).
II. There exists a continuous linear functional \(G\) on \((\ell^{2}, \|\cdot\|_{2})\) such that \(G = f\) on \(C_{00}\).

Which one of the following is correct?

A.Both I and II are TRUE
B.I is TRUE and II is FALSE
C.I is FALSE and II is TRUE
D.Both I and II are FALSE

Exam Year(s)

GATE Mathematics - 2024,

Answer: B

I is TRUE and II is FALSE.

I: for \(x \in C_{00}\), \(|f(x)| \le \sum|x_{n}|/\sqrt{n} \le \|x\|_{1}\), so \(f\) is bounded in the \(\ell^{1}\) norm. Since \(C_{00}\) is dense in \(\ell^{1}\), \(f\) extends to a continuous linear functional \(F\) on \(\ell^{1}\) (given by the bounded sequence \(1/\sqrt{n}\)).

II: a continuous extension to \(\ell^{2}\) would be \(G(x) = \sum x_{n}c_{n}\) with \((c_{n}) \in \ell^{2}\), forcing \(c_{n} = 1/\sqrt{n}\). But \(\sum 1/n\) diverges, so \((1/\sqrt{n}) \notin \ell^{2}\). Concretely, \(x = (1/\sqrt{n})_{n \le N}\) has \(\|x\|_{2}^{2} = f(x) = \sum_{n \le N}1/n\), so \(f(x)/\|x\|_{2}\) is unbounded and \(f\) is not continuous on \(\ell^{2}\).

Discuss in Forum

Let \(X = \{f \in C[0,1] : f(0) = 0 = f(1)\}\) with the norm \(\|f\|_{\infty} = \sup_{0 \le t \le 1} |f(t)|\), where \(C[0,1]\) is the space of all real-valued continuous functions on [0, 1]. Let \(Y = C[0,1]\) with the norm \(\|f\|_{2} = \left(\int_{0}^{1} |f(t)|^{2} dt\right)^{1/2}\). Let \(U_{X}\) and \(U_{Y}\) be the closed unit balls in X and Y centred at the origin, respectively. Consider \(T: X \to \mathbb{R}\) and \(S: Y \to \mathbb{R}\) given by

\[ Tf = \int_{0}^{1} f(t)\,dt \]

and

\[ Sf = \int_{0}^{1} f(t)\,dt. \]

Consider the following statements:

S1: \(\sup_{f \in U_{X}} |Tf|\) is attained at a point of \(U_{X}\).
S2: \(\sup_{f \in U_{Y}} |Sf|\) is attained at a point of \(U_{Y}\).

Then, which one of the following is correct?

A.S1 is TRUE and S2 is FALSE
B.S2 is TRUE and S1 is FALSE
C.both S1 and S2 are TRUE
D.neither S1 nor S2 is TRUE

Exam Year(s)

GATE Mathematics - 2025,

Answer: B

S2 is TRUE and S1 is FALSE.

S1: for f in \(U_{X}\), \(|Tf| \le \int_{0}^{1}|f| \le 1\), and functions equal to 1 except near the end points make \(Tf\) as close to 1 as we like, so the supremum is 1. But reaching 1 would need \(|f(t)| = 1\) everywhere, impossible for a continuous f with \(f(0) = f(1) = 0\). The supremum is not attained.

S2: by the Cauchy-Schwarz inequality, \(|Sf| = |\int_{0}^{1} f \cdot 1| \le \|f\|_{2}\|1\|_{2} \le 1\), with equality for the constant function f = 1, which lies in \(U_{Y}\) since its norm is 1. The supremum 1 is attained. So only S2 holds.

Discuss in Forum

Consider the following two spaces:

\(X = (C[-1,1], \|\cdot\|_{\infty})\), the space of all real-valued continuous functions defined on [-1, 1] equipped with the norm \(\|f\|_{\infty} = \sup_{t \in [-1,1]} |f(t)|\).

\(Y = (C[-1,1], \|\cdot\|_{2})\), the space of all real-valued continuous functions defined on [-1, 1] equipped with the norm \(\|f\|_{2} = \left(\int_{-1}^{1} |f(t)|^{2} dt\right)^{1/2}\).

Let W be the linear span over \(\mathbb{R}\) of all the Legendre polynomials. Then, which one of the following is correct?

A.W is dense in X but not in Y
B.W is dense in Y but not in X
C.W is dense in both X and Y
D.W is dense neither in X nor in Y

Exam Year(s)

GATE Mathematics - 2025,

Answer: C

W is dense in both X and Y.

The Legendre polynomials \(P_{0}, P_{1}, P_{2}, \ldots\) have degrees 0, 1, 2, ..., so their span W is the set of all polynomials. By the Weierstrass approximation theorem, polynomials are dense in \(C[-1, 1]\) under the sup norm, so W is dense in X.

For Y, note that \(\|f\|_{2} \le \sqrt{2}\|f\|_{\infty}\) on [-1, 1]. Any polynomial within \(\varepsilon\) of f in the sup norm is within \(\sqrt{2}\varepsilon\) in the 2-norm, so density in X carries over to Y. The weaker norm can only make it easier to approximate, never harder.

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Let X be the space of all continuously differentiable real valued functions on [0,1]. Define the following norms on X:

\[ p_1(x) = \sup\{|x(t)| : t \in [0,1]\} \]

\[ p_2(x) = \sup\{|\tfrac{d}{dt} x(t)| : t \in [0,1]\} \]

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.

Exam Year(s)

GATE Mathematics - 2026,

Answer: D

The true statement is \((X, p_3)\) is a Banach space.

\(p_3\) is the usual \(C^1\) norm. If \((x_n)\) is Cauchy in \(p_3\), both \(x_n\) and \(x_n'\) converge uniformly, to some x and g; uniform convergence of the derivatives lets you pass to the limit in \(x_n(t) = x_n(0) + \int_0^t x_n'\), so x is continuously differentiable with \(x' = g\), and \(x_n \to x\) in \(p_3\). The space is complete.

\(p_1\) alone fails: smooth functions can converge uniformly to \(|t - \frac{1}{2}|\), which is not differentiable, so \((X, p_1)\) is not complete. \(p_2\) vanishes on every constant function, so it is not even a norm and \((X, p_2)\) is not a Banach space; the option saying \(p_3\) fails is the negation of the true statement.

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10.

Let

\[ \begin{aligned} \ell^{\infty} = \{ & x = (x_n)_{n \geq 1} \mid x_n \in \mathbb{R}, \\ & \sup\{|x_n| : n = 1, 2, \ldots\} \lt \infty\} \end{aligned} \]

with the supremum norm. Let \(T : \ell^{\infty} \to \ell^{\infty}\) be given by \(T(x_1, x_2, x_3, \ldots) = (x_1, \frac{x_2}{2}, \frac{x_3}{3}, \ldots)\).

Which one of the following is TRUE?

A.T is bounded but not one-to-one.
B.T is one-to-one but not bounded.
C.T is bounded and the inverse (from the range of T) exists but not bounded.
D.T is bounded and the inverse (from the range of T) is bounded.

Exam Year(s)

GATE Mathematics - 2026,

Answer: C

The true statement is T is bounded and the inverse (from the range of T) exists but not bounded.

T is bounded: \(|x_n / n| \le |x_n|\), so \(\|Tx\|_{\infty} \le \|x\|_{\infty}\) and \(\|T\| = 1\). It is one-to-one: \(Tx = 0\) forces every \(x_n = 0\). So the inverse exists on the range, and it is \((y_n) \mapsto (n y_n)\).

That inverse is unbounded. The unit vector \(e_n\) is mapped by T to \(e_n / n\), of norm \(1/n\), and the inverse sends this back to \(e_n\), of norm 1, so

\[ \frac{\|T^{-1}(e_n/n)\|}{\|e_n/n\|} = n \to \infty. \]

So T is not 'not one-to-one', is not unbounded, and does not have a bounded inverse.

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