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.
Answer the question set and review your mistakes at the end.
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 \(0\).
