Try
A sequence can be small in several different senses. Consider x=(1/2,1/4,1/8,…). Its total absolute size is 1, its sum of squares is 1/3, and its largest coordinate magnitude is 1/2. The square-root-of-squares size is therefore 1/√3. These are three different norms of the same sequence, answering different questions.
Compare x with y=(1,1/√2,1/√3,…). The coordinates of y tend to zero and stay bounded by 1, yet their squares sum to the divergent harmonic series. Coordinate decay alone does not guarantee a finite sum of squares.
| Sequence indexed by n≥1 | Sum of magnitudes | Square-root of sum of squares | Supremum magnitude | Coordinates tend to zero? |
|---|
| 2^(−n) | 1 | 1/√3 | 1/2 | Yes |
| n^(−1/2) | Diverges | Diverges | 1 | Yes |
| 1 | Diverges | Diverges | 1 | No |
Predict whether seeing the first hundred small coordinates would decide any of these infinite-series questions. It cannot without a rule and an argument controlling the remaining terms. The formulas, not just the displayed samples, allow the conclusions.
Predict
For real p with 1≤p<∞, define ℓᵖ to consist of real sequences x=(xₙ) with Σ|xₙ|ᵖ finite, and define ||x||ₚ=(Σ|xₙ|ᵖ)^(1/p). The nonnegative series means the increasing limit of finite partial sums. Define ℓ∞ to consist of bounded real sequences, with ||x||∞=sup over n of |xₙ|. This supremum need not be attained: the sequence 1−1/n has supremum 1 but no term equal to 1.
Define c₀ as the sequences whose coordinates tend to zero, and c₀₀ as the sequences with only finitely many nonzero coordinates. Both sit inside ℓ∞ and can use its supremum norm. Finite-support sequences also belong to every finite ℓᵖ and may carry those different norms.
Predict the inclusion direction when 1≤p<q<∞. For summable sequences, the small tail entries shrink when raised to a larger power, so ℓᵖ will be contained in ℓᑫ. Do not transfer an intuition from integrals over a finite-measure interval: the actual sequence proof must determine the direction.
Experiment
Write e⁽ᵏ⁾ for a sequence whose only nonzero entry is a 1 in coordinate k. Display the first four such sequences as rows. Follow one fixed column downward: it becomes zero eventually. Now compute each row’s full norm. Every row has norm 1, for every finite p and for the supremum norm.
This is a moving-coordinate experiment: e⁽ᵏ⁾ tends coordinatewise to zero but does not tend to zero in norm. Distinct rows differ in two coordinates, so their distance is 2^(1/p) for finite p and 1 for the supremum norm. Even selecting a subsequence cannot make them norm Cauchy.
For a second paper action, use z⁽ᴺ⁾ with its first N coordinates equal to 1 and the rest zero. Its p-norm is N^(1/p), while its supremum norm is 1. At N=16, the 1-, 2-, 4- and supremum norms are 16, 4, 2 and 1. Increasing the support reveals why a norm-equivalence constant valid in one finite dimension need not work across all sequences.
Understand
Finite Hölder needs only a scalar inequality
Let 1<p<∞ and r=p/(p−1), so 1/p+1/r=1. For a,b≥0, Young’s inequality states ab≤aᵖ/p+bʳ/r. Here is a local proof without a measure-theory prerequisite. For fixed b>0, the function φ(a)=aᵖ/p−ab+bʳ/r has derivative a^(p−1)−b. It decreases up to a=b^(1/(p−1)) and increases afterward; substitution at that minimum gives zero. The boundary a=0 is nonnegative, and b=0 is immediate.
For finite vectors with positive norms A=(Σ|xᵢ|ᵖ)^(1/p) and B=(Σ|yᵢ|ʳ)^(1/r), apply Young to |xᵢ|/A and |yᵢ|/B and sum. The two power sums are both 1, so Σ|xᵢyᵢ|≤AB. If a norm is zero, its vector is zero and the inequality holds directly. For p=1 use the separate estimate Σ|xᵢyᵢ|≤(Σ|xᵢ|)max|yᵢ|; the conjugate-exponent formula is not used at this endpoint.
To obtain infinite Hölder, apply the finite inequality to each prefix of x∈ℓᵖ and y∈ℓʳ. Every partial sum of |xₙyₙ| is bounded by ||x||ₚ||y||ᵣ. These nonnegative partial sums are increasing and bounded, so their real limit exists and has the same bound. For x∈ℓ¹,y∈ℓ∞ the endpoint estimate gives the identical conclusion with ||y||∞. This is a controlled passage through partial sums, not an unexplained substitution of infinity into a finite formula.
Minkowski establishes both addition and the norm inequality
On a finite prefix put C=(Σ|xᵢ+yᵢ|ᵖ)^(1/p), with 1<p<∞. If C=0 the desired inequality is automatic. Otherwise write
Σ|xᵢ+yᵢ|ᵖ ≤ Σ|xᵢ||xᵢ+yᵢ|^(p−1) + Σ|yᵢ||xᵢ+yᵢ|^(p−1).
Apply finite Hölder to both sums. The r-norm of the factor |xᵢ+yᵢ|^(p−1) is C^(p−1). Therefore Cᵖ≤(A+B)C^(p−1), giving C≤A+B. For infinite x,y∈ℓᵖ, this bounds every partial p-power sum of x+y by (||x||ₚ+||y||ₚ)ᵖ. Increasing bounded partial sums converge; hence x+y belongs to ℓᵖ, and their limits give the triangle inequality. We proved membership rather than assuming it in order to write its norm.
For p=1, sum the ordinary absolute-value triangle inequality. For ℓ∞, bound each |xₙ+yₙ| by ||x||∞+||y||∞ and take a supremum. Nonnegativity, definiteness and absolute homogeneity follow directly: a zero norm forces every coordinate to be zero, and multiplying by a scalar multiplies the norm by its absolute value. These arguments also establish scalar closure and the vector-space norm axioms.
Inclusions have quantitative proofs
Suppose x∈ℓᵖ and A=||x||ₚ>0. Each |xₙ|≤A, because its pth power is one term of the total nonnegative sum. Thus uₙ=|xₙ|/A lies in [0,1]. For q>p, uₙᑫ≤uₙᵖ, so Σuₙᑫ≤1. It follows that x∈ℓᑫ and ||x||ᑫ≤||x||ₚ. The zero case is immediate. The same single-coordinate bound gives ||x||∞≤||x||ₚ.
If x∈ℓᑫ did not tend to zero, some ε>0 would satisfy |xₙ|≥ε at infinitely many indices. Its partial q-power sums would then exceed arbitrarily large multiples of εᑫ, contradicting summability. Hence ℓᑫ⊆c₀. A sequence tending to zero is bounded: its late entries are at most 1 in magnitude, and its finite prefix has a finite bound. Thus c₀⊆ℓ∞. Sums and scalar multiples of null sequences remain null by epsilon estimates, so c₀ is a subspace. Finite-support sequences plainly have finite power sums.
Apply
Prove every inclusion is strict when 1≤p<q<∞. Set a=(1/p+1/q)/2. Then ap<1 but aq>1, so xₙ=n^(−a) belongs to ℓᑫ but not ℓᵖ. To justify the power-decay series test used here, group terms with 2ᵏ≤n<2^(k+1). For exponent s>1 the block is at most 2^(k(1−s)), and these bounds form a convergent geometric series. For 0<s≤1, n^(−s)≥1/n, while every harmonic block contributes at least 1/2. Thus the series diverges.
Next, n^(−1/q) is in c₀ but not ℓᑫ because its q-power sum is harmonic. The constant sequence 1 belongs to ℓ∞ but not c₀. These witnesses prove
ℓᵖ ⊊ ℓᑫ ⊊ c₀ ⊊ ℓ∞.
The geometric sequence 2^(−n) has infinite support but finite p-power sum for every p≥1, so c₀₀ is properly contained in every finite ℓᵖ too.
Return to the block vectors z⁽ᴺ⁾. For p<q their norm ratio is N^(1/p−1/q), interpreting 1/∞ as zero when q=∞. It grows without bound. Therefore no constant C can make ||z||ₚ≤C||z||ᑫ for every finite-support z. This does not deny finite-dimensional norm equivalence; it shows why a constant that depends on dimension cannot prove an infinite-dimensional statement.
Continue
Record which convergence your argument uses. Coordinatewise convergence fixes an index before taking a limit across sequences. Norm convergence requires a single bound on the total sum or supremum across every coordinate. The moving unit vectors show that one cannot replace the other.
The next lesson will prove completeness and analyze truncation errors. Today’s norm axioms and inclusion inequalities do not already establish completeness, density of finite-support sequences, or compactness of bounded sets. Those require separate arguments about limits and tails. The finite Hölder proof above supplies the necessary local inequality without requiring the measure-theoretic Lᵖ course first.
