Learn Normed Vector Spaces: Interactive Problems and Worked Examples
Use the question set below to practice normed vector spaces: norm axioms, \(d(x,y)=\|x-y\|\), open and closed balls, \(\ell^1\), Euclidean, and \(\ell^\infty\) norms, norm convergence, Cauchy sequences, Banach spaces, finite-dimensional norm equivalence, continuity of the norm map, addition and scalar multiplication of convergent sequences, normalization \(\frac{x}{\|x\|}\), and basic operator norms such as the identity and zero map. If you want a refresher, open the lesson for mentally followable examples and checks.
Jawab rangkaian soal dan tinjau kesalahanmu di akhir.
How this normed vector spaces practice works
1. Take the practice set: answer norm axiom, convergence, completeness, and operator norm questions below.
2. Open the lesson: review norm properties, standard examples, norm-induced topology, Banach spaces, and finite-dimensional shortcuts.
3. Retry: return to the question set and use norm language immediately.
What you will learn in the normed vector spaces lesson
Norm axioms and distance
Positive definiteness: \(\|x\|=0\) exactly when \(x=0\)
Purpose: Build a working picture of a normed vector space: a vector space with a length function \(\|x\|\). From that length you get distance, convergence, balls, continuity, completeness, and a way to measure linear maps.
Success criteria
Check the three norm axioms and use \(\|x\|=0\Rightarrow x=0\).
Compute \(\ell^1\), Euclidean, and \(\ell^\infty\) norms in small coordinate examples.
Use the induced metric \(d(x,y)=\|x-y\|\).
Describe open balls, closed balls, and the unit sphere.
Translate \(x_n\to x\) into \(\|x_n-x\|\to0\), and use \(x_n+y_n\to x+y\), \(ax_n\to ax\), and \(\|x_n\|\to\|x\|\).
Normalize a nonzero vector: \(\left\|\frac{x}{\|x\|}\right\|=1\).
Explain Cauchy sequences and Banach spaces.
Use finite-dimensional norm equivalence to transfer convergence and continuity facts.
Compute simple operator norms for the zero and identity maps.
Key vocabulary
Norm: a function \(\|\cdot\|:V\to[0,\infty)\) satisfying positivity, homogeneity, and the triangle inequality.
Norm-induced metric: \(d(x,y)=\|x-y\|\).
Open ball: \(B(a,r)=\{x:\|x-a\|\lt r\}\).
Cauchy sequence: a sequence whose terms eventually become arbitrarily close to each other.
Banach space: a complete normed vector space.
Operator norm: \(\|T\|=\sup_{\|x\|\le1}\|Tx\|\) for a bounded linear map.
Quick pre-check
Pre-check 1: In a normed vector space, what does \(\|x\|=0\) imply?
Hint: Positive definiteness says only the zero vector has norm zero.
Pre-check 2: Which formula defines the distance induced by a norm?
Hint: Distance is measured by the norm of the difference.
A norm is length compatible with vector structure
Learning goal: Recognize the norm axioms quickly and use the triangle inequality to derive common estimates.
Key idea
A norm on a real or complex vector space \(V\) satisfies \(\|x\|\ge0\), \(\|x\|=0\) exactly when \(x=0\), \(\|ax\|=|a|\|x\|\), and \(\|x+y\|\le\|x\|+\|y\|\). Thus \(x≠0\) gives \(\|x\|>0\), and the normalized vector \(\frac{x}{\|x\|}\) has norm \(1\). The scalar absolute value is essential: negative scalars cannot make length negative.
Recognition checklist
Zero test: if a proposed norm gives \(0\) to a nonzero vector, it fails.
Scalar test: check \(\|ax\|=|a|\|x\|\), especially for \(a=-1\).
Triangle test: check \(\|x+y\|\le\|x\|+\|y\|\); for example, \(\|x\|,\|y\|\le1\) gives \(\|x+y\|\le2\).
Reverse triangle: \(|\|x\|-\|y\||\le\|x-y\|\) follows from the triangle inequality.
Worked example
Example: If \(\|x\|=3\), what is \(\|-2x\|\)?
Use homogeneity: \(\|-2x\|=|-2|\,\|x\|=2\cdot3=6\). The sign of the scalar disappears because length is nonnegative.
Try it
Try it 1: For a scalar \(a\), which identity must a norm satisfy?
Hint: Pull out the absolute value of the scalar.
Try it 2: Which inequality is the reverse triangle inequality?
Hint: Compare the two lengths \(\|x\|\) and \(\|y\|\) by measuring \(x-y\).
Three coordinate norms to know on sight
Learning goal: Compute \(\ell^1\), Euclidean, and \(\ell^\infty\) norms and recognize their unit balls.
In \(\mathbb{R}^2\), the \(\ell^1\) unit ball is a diamond, the Euclidean unit ball is a disk, and the \(\ell^\infty\) unit ball is a square.
Worked example
Example: Compute the three standard norms of \(v=(3,-4)\).
\(\|v\|_1=3+4=7\), \(\|v\|_2=\sqrt{3^2+(-4)^2}=5\), and \(\|v\|_\infty=\max(3,4)=4\). Each norm measures the same vector with a different geometry.
Try it
Try it 1: What is \(\|(3,-4)\|_1\)?
Hint: The \(\ell^1\) norm adds absolute coordinate values.
Try it 2: In \(\mathbb{R}^2\), which norm has a square-shaped unit ball?
Hint: \(\|x\|_\infty\le1\) means every coordinate lies between \(-1\) and \(1\).
The norm turns vectors into a metric space
Learning goal: Use norm-induced balls and convergence statements without confusing strict and non-strict inequalities.
Key idea
A norm defines \(d(x,y)=\|x-y\|\), so small \(\|x-y\|\) means \(x\) and \(y\) are close. Then \(x_n\to x\) means \(\|x_n-x\|\to0\). Limits behave naturally with vector operations: if \(x_n\to x\), \(y_n\to y\), and \(a\) is fixed, then \(x_n+y_n\to x+y\) and \(ax_n\to ax\). The open ball centered at \(a\) with radius \(r\) is \(B(a,r)=\{x:\|x-a\|\lt r\}\); the closed ball uses \(\le r\); the unit sphere uses \(=1\).
Worked example
Example: If \(\|x_n-x\|\le1/n\), prove \(x_n\to x\).
Since \(0\le\|x_n-x\|\le1/n\) and \(1/n\to0\), the squeeze principle gives \(\|x_n-x\|\to0\). By definition, \(x_n\to x\) in norm.
Try it
Try it 1: In a normed vector space, what is the open ball \(B(a,r)\)?
Hint: Open means strictly less than the radius, measured by \(\|x-a\|\).
Try it 2: If \(\|x_n-x\|\le1/n\), then \(x_n\):
Hint: The upper bound tends to \(0\).
Completeness means Cauchy sequences have limits inside
Learning goal: Separate convergence, Cauchy behavior, and completeness.
Key idea
Every convergent sequence is Cauchy, but a Cauchy sequence might fail to converge inside the space. A normed vector space is complete when every Cauchy sequence has a limit in the space. A complete normed vector space is called a Banach space.
Worked example
Example: Why is completeness a property of the space, not just of a sequence?
The Cauchy condition only says terms get close to each other. Completeness asks whether every such sequence has a limit that still belongs to the space. A missing limit point makes the space incomplete even though the sequence is Cauchy.
Try it
Try it 1: A Banach space is a normed vector space that is:
Hint: Banach means complete for the metric induced by the norm.
Try it 2: A sequence that converges in a normed space is always:
Hint: Once a sequence is close to its limit, two late terms are close to each other.
Finite dimension makes all norms topologically alike
Learning goal: Use norm equivalence in finite-dimensional spaces and avoid extending it blindly to infinite-dimensional spaces.
Key idea
Two norms \(\|\cdot\|_a\) and \(\|\cdot\|_b\) are equivalent if there are constants \(m,M>0\) such that \(m\|x\|_a\le\|x\|_b\le M\|x\|_a\) for all \(x\). In finite dimension, all norms are equivalent, so they define the same open sets and the same convergent sequences.
Worked example
Example: Compare \(\|x\|_\infty\), \(\|x\|_2\), and \(\|x\|_1\) on \(\mathbb{R}^2\).
For \(x=(a,b)\), \(\|x\|_\infty\le\|x\|_2\le\|x\|_1\le2\|x\|_\infty\). These inequalities show the norms control each other, so convergence in one is convergence in the others.
Try it
Try it 1: In a finite-dimensional real vector space, two norms are:
Hint: Finite dimension is the key phrase.
Try it 2: If two norms on a finite-dimensional space are equivalent, they define the same:
Hint: Equivalent norms have the same open sets and convergent sequences.
For a linear map \(T:V\to W\), boundedness means \(\|Tx\|\le C\|x\|\) for some \(C\). This is equivalent to continuity for linear maps. The operator norm is \(\|T\|=\sup_{\|x\|\le1}\|Tx\|\). Between finite-dimensional normed spaces, every linear map is automatically continuous and uniformly continuous.
Worked example
Example: What are the operator norms of the zero map and the identity map on a nonzero normed space?
The zero map sends every vector to \(0\), so its operator norm is \(0\). The identity map preserves every vector, so \(\|Ix\|=\|x\|\). On a nonzero normed space, vectors of norm \(1\) exist, so the identity operator norm is \(1\).
Try it
Try it 1: The zero map \(x\mapsto0\) has operator norm:
Hint: Every unit vector is sent to the zero vector.
Try it 2: Every linear map between finite-dimensional normed spaces is:
Hint: In finite dimension, linear maps are Lipschitz after choosing any norm.
Avoid the common normed-space mix-ups
Learning goal: Finish with common traps and a compact method for quiz problems.
Common traps
A norm is not usually linear: \(\|x+y\|\) is usually not \(\|x\|+\|y\|\).
Homogeneity needs absolute value: \(\|-x\|=\|x\|\), not \(-\|x\|\).
Open ball vs sphere: \(\lt r\) is a ball, \(=r\) is a sphere.
Cauchy is not the same as convergent unless the space is complete.
Finite-dimensional norm equivalence does not automatically cover infinite-dimensional spaces.
Operator norm uses a supremum over the unit ball, not a value at one convenient vector.
Worked example
Example: If \(\|x-y\|=0\), what follows?
By positive definiteness, \(\|x-y\|=0\) implies \(x-y=0\). Therefore \(x=y\). This is exactly why \(d(x,y)=\|x-y\|\) separates points.
Try it
Try it 1: Is the map \(x\mapsto\|x\|\) generally linear?
Hint: Test scalar multiplication with a negative scalar.
Try it 2: If \(\|x-y\|=0\), then:
Hint: Apply the zero-norm property to the vector \(x-y\).
Final recap
A norm has positivity, homogeneity with \(|a|\), and the triangle inequality.
Every norm gives a metric by \(d(x,y)=\|x-y\|\).
Norm convergence is \(\|x_n-x\|\to0\), and it is compatible with addition, fixed scalar multiplication, and the norm map.
If \(x≠0\), then \(\|x\|>0\) and \(\left\|\frac{x}{\|x\|}\right\|=1\).
A Banach space is a complete normed vector space.
All norms on a finite-dimensional vector space are equivalent and give the same topology.
Linear maps between finite-dimensional normed spaces are automatically continuous.
The zero map has operator norm \(0\), and the identity map on a nonzero normed space has operator norm \(1\).
Next step: Close this lesson and try the quiz again. When a question mentions zero norm, use positive definiteness; when it mentions convergence, translate it into a norm going to \(0\).
