Convergence in a Norm

Convergence Measured by a Norm

The Metric Induced by a Norm showed that a norm measures distances by $d_N(x,y)=N(x-y)$. This makes convergence in a normed space a particular case of convergence in a metric space: the distance from each term to the proposed limit must become arbitrarily small. The norm notation is especially useful because it also interacts directly with vector addition and scalar multiplication.

Let $V$ be a real vector space with norm $N$, and let $(x_n)$ be a sequence in $V$. The proposed limit must itself belong to $V$. We write $x_n\to x$ in norm when the norm of the difference $x_n-x$ tends to zero.

Definition: A sequence $(x_n)$ in a normed space $(V,N)$ converges in norm to $x\in V$ if, for every $\varepsilon>0$, there is an integer $n_0$ such that $$ n\geq n_0\quad\Longrightarrow\quad N(x_n-x)<\varepsilon. $$ We write $x_n\to x$, or $N(x_n-x)\to0$. Since $d_N(x_n,x)=N(x_n-x)$, this is exactly convergence in the metric induced by $N$.

The definition says that eventually all terms lie in every open ball centered at $x$, no matter how small the radius. The index $n_0$ may depend on $\varepsilon$, but not on $n$ once $n\geq n_0$. Convergence is therefore a statement about the tail of the sequence; the first finitely many terms do not affect whether the sequence converges.

First Examples in Normed Spaces

Worked Example: A Sequence in the Plane

In $\mathbb{R}^2$, let $x_n=(1/n,(-1)^n/n)$ and $x=(0,0)$. For the Euclidean norm, $$ \|x_n-x\|_2 =\sqrt{\left(\frac1n\right)^2+\left(\frac{(-1)^n}{n}\right)^2} =\frac{\sqrt{2}}{n}, $$ because $((-1)^n)^2=1$. Given $\varepsilon>0$, choose an integer $n_0>\sqrt{2}/\varepsilon$. Then $n\geq n_0$ implies $\|x_n-x\|_2=\sqrt{2}/n<\varepsilon$, so $x_n\to(0,0)$ in the Euclidean norm.

The same sequence converges in the other two standard coordinate norms. In the $1$-norm, $\|x_n\|_1=1/n+|(-1)^n|/n=2/n$. In the supremum norm, $\|x_n\|_\infty=1/n$. Each quantity tends to zero. These calculations illustrate a useful finite-dimensional fact: the Equivalence of Norms in Finite Dimensions, established earlier in the course, ensures that all norms on $\mathbb{R}^2$ give the same convergent sequences.

Worked Example: Convergence of Polynomials in a Coefficient Norm

On $\mathcal{P}_2$, use the norm $N(a_0+a_1x+a_2x^2)=|a_0|+2|a_1|+|a_2|$. Define $$ p_n(x)=\left(1+\frac1n\right)+\left(2-\frac1n\right)x+\left(1+\frac2n\right)x^2 $$ and $p(x)=1+2x+x^2$. Subtracting coefficients gives $$ p_n-p=\frac1n-\frac1n x+\frac2n x^2, \qquad N(p_n-p)=\frac1n+2\left|\!-\frac1n\right|+\frac2n=\frac5n. $$ For every $\varepsilon>0$, choose $n_0>5/\varepsilon$. For $n\geq n_0$, $N(p_n-p)=5/n<\varepsilon$. Thus $p_n\to p$ in this norm. The calculation uses the norm of the difference polynomial, rather than comparing the separate norms of $p_n$ and $p$.

Worked Example: Convergence Depends on the Function Norm

Consider $f_n(x)=x^n$ on $[0,1]$, with the zero function as the proposed limit. In the $L^1$ norm on $C[0,1]$, $$ \|f_n-0\|_1=\int_0^1x^n\,dx=\frac{1}{n+1}\longrightarrow0. $$ Indeed, for any $\varepsilon>0$, choosing $n_0$ so that $1/(n_0+1)<\varepsilon$ ensures $1/(n+1)<\varepsilon$ for all $n\geq n_0$. Hence $f_n\to0$ in the $L^1$ norm.

In the supremum norm, however, $\|f_n-0\|_\infty=\sup_{0\leq x\leq1}x^n=1$, since $x^n\leq1$ throughout the interval and $1^n=1$. This quantity does not tend to zero. Therefore $f_n$ does not converge to zero in the supremum norm. There is no contradiction: convergence is measured using the chosen norm. In particular, the values at the single endpoint $x=1$ prevent the supremum error from becoming small, while the $L^1$ error is the integral $1/(n+1)$.

Uniqueness of a Norm Limit

Theorem (Uniqueness of Norm Limits): If a sequence in a normed space converges in norm to both $x$ and $y$, then $x=y$.

Proof. Suppose $x_n\to x$ and $x_n\to y$. By the triangle inequality, $$ N(x-y)=N\bigl((x-x_n)+(x_n-y)\bigr) \leq N(x-x_n)+N(x_n-y). $$ Let $\varepsilon>0$. Since $x_n\to x$, for all sufficiently large $n$, $N(x-x_n)<\varepsilon/2$. Since $x_n\to y$, for all sufficiently large $n$, $N(x_n-y)<\varepsilon/2$. Taking $n$ large enough for both inequalities to hold gives $N(x-y)<\varepsilon$. This is true for every $\varepsilon>0$, so $N(x-y)=0$. Positive definiteness of the norm implies $x-y=0$, and hence $x=y$. $\square$

This result is also the uniqueness of limits for the induced metric, expressed using the norm. It is important when a limit is identified by two different methods: the conclusions must agree. Uniqueness does not say that every sequence has a limit. It says only that there cannot be two distinct limits in the space.

Limit Laws for Vector Operations

The norm axioms make addition and scalar multiplication compatible with convergence. The next theorem is a practical way to combine sequences whose limits are already known.

Theorem (Limit Laws for Norm Convergence): Suppose $x_n\to x$ and $y_n\to y$ in a normed space $V$. For every fixed scalar $c\in\mathbb{R}$, $$ x_n+y_n\to x+y,\qquad cx_n\to cx. $$

Proof. For the sum, the triangle inequality gives $$ N\bigl((x_n+y_n)-(x+y)\bigr) =N\bigl((x_n-x)+(y_n-y)\bigr) \leq N(x_n-x)+N(y_n-y). $$ Let $\varepsilon>0$. Convergence of $x_n$ to $x$ gives an index after which $N(x_n-x)<\varepsilon/2$; convergence of $y_n$ to $y$ gives an index after which $N(y_n-y)<\varepsilon/2$. Beyond the larger of these two indices, the displayed sum is less than $\varepsilon$. Thus $x_n+y_n\to x+y$.

For scalar multiplication, if $c=0$, then $cx_n=cx=0$ for every $n$, so convergence is immediate. If $c\neq0$, homogeneity gives $$ N(cx_n-cx)=|c|N(x_n-x). $$ Given $\varepsilon>0$, convergence of $x_n$ provides an index after which $N(x_n-x)<\varepsilon/|c|$. For all such $n$, $N(cx_n-cx)<\varepsilon$. Therefore $cx_n\to cx$. $\square$

By applying these two laws repeatedly, any fixed finite linear combination of convergent sequences converges to the same linear combination of their limits. The coefficients must be fixed: this theorem does not make a claim about multiplying $x_n$ by a scalar that changes with $n$. In a general normed space, there may also be no multiplication operation between two vectors.

Worked Example: Combining Two Convergent Sequences

In $\mathbb{R}^2$ with the Euclidean norm, let $$ u_n=\left(1+\frac1n,\frac2n\right),\qquad v_n=\left(-1,\ 3-\frac1n\right). $$ The differences from their proposed limits $u=(1,0)$ and $v=(-1,3)$ have norms $$ \|u_n-u\|_2=\sqrt{\frac1{n^2}+\frac4{n^2}}=\frac{\sqrt5}{n}, \qquad \|v_n-v\|_2=\frac1n. $$ Thus $u_n\to u$ and $v_n\to v$. The limit laws give $u_n+2v_n\to u+2v$, where $$ u_n+2v_n=\left(-1+\frac1n,\ 6\right), \qquad u+2v=(-1,6). $$ Directly, the difference is $(1/n,0)$, whose Euclidean norm is $1/n\to0$, confirming the conclusion.

The Norm of a Convergent Sequence

Convergence controls not only the distance to the limit, but also the sizes of the terms. The Reverse Triangle Inequality for Norms, proved earlier in Norm Axioms, states that $|N(u)-N(v)|\leq N(u-v)$. Applying it to a term and its limit gives the result below.

Theorem (Continuity of the Norm): If $x_n\to x$ in a normed space, then the real sequence $N(x_n)$ converges to $N(x)$.

Proof. The Reverse Triangle Inequality gives, for every $n$, $$ |N(x_n)-N(x)|\leq N(x_n-x). $$ Given $\varepsilon>0$, norm convergence supplies an index $n_0$ such that $N(x_n-x)<\varepsilon$ whenever $n\geq n_0$. The displayed inequality then yields $|N(x_n)-N(x)|<\varepsilon$ for all $n\geq n_0$. This is precisely $N(x_n)\to N(x)$. $\square$

For example, if $x_n\to x$, the theorem guarantees $N(x_n)\to N(x)$, even when calculating each norm separately would be difficult. The converse is not true: convergence of the numbers $N(x_n)$ alone does not show that the vectors approach a particular vector. A sequence can keep the same norm while changing direction.

Using the Definition Carefully

A direct proof of norm convergence usually follows a short sequence of decisions: identify the proposed limit, calculate or estimate the norm of the error, and choose an index that makes the bound smaller than the requested tolerance. The estimate must hold for every sufficiently large index, not just for selected terms.

Compute the error.
Write $x_n-x$ explicitly and evaluate or bound $N(x_n-x)$.

Set the target.
Given $\varepsilon>0$, find an inequality that ensures $N(x_n-x)<\varepsilon$.

Choose an index.
Select $n_0$ so the inequality holds whenever $n\geq n_0$, and verify the resulting bound.

A common mistake is to switch silently between different norms. The function sequence $x^n$ illustrates why this matters: it converges to zero in the $L^1$ norm but not in the supremum norm. Another mistake is to confuse pointwise convergence with norm convergence. Pointwise convergence tests the values at each fixed input separately; norm convergence tests the size of the entire difference according to one specified norm. The two notions can agree in some settings, but neither should be substituted for the other without a result that justifies it.

Finally, the definition requires a limit inside $V$. A sequence can have errors that become small in a natural sense while its would-be limit is not an element of the space. Whether every sequence satisfying an appropriate Cauchy condition actually converges in the space is a separate question; it depends on completeness and is the subject of the next tutorial.

Takeaway: Norm convergence means $N(x_n-x)\to0$. It is exactly convergence in the induced metric, has a unique limit, respects addition and fixed scalar multiplication, and forces $N(x_n)\to N(x)$. The chosen norm matters.

Check Your Understanding

Use the definition and results in this tutorial to answer the following questions.

What must be shown, for every $\varepsilon>0$, to prove that $x_n\to x$ in norm?
Why does the triangle inequality imply that a sequence cannot have two distinct norm limits?
If $x_n\to x$ and $y_n\to y$, what is the limit of $3x_n-y_n$, and which limit laws justify it?
If $x_n\to x$, what can be concluded about the real sequence $N(x_n)$? Which earlier inequality supports the conclusion?
For $f_n(t)=t^n$ on $[0,1]$, what are $\|f_n\|_1$ and $\|f_n\|_\infty$, and what do they show about dependence on the norm?

What You'll Learn