16.3: Convergence of Sequences of Vectors

Convergence of Vectors

We now discuss pointwise and norm convergence of vectors. Other types of convergence also exist, and one in particular, uniform convergence (Section 16.4), can also be studied. For this discussion , we will assume that the vectors belong to a normed vector space (Section 15.3).

Pointwise Convergence

A sequence (Section 16.2) \(\left.\left\{g_{n}\right\}\right|_{n=1} ^{\infty}\) converges pointwise to the limit \(\boldsymbol{g}\) if each element of \(g_n\) converges to the corresponding element in \(\boldsymbol{g}\). Below are few examples to try and help illustrate this idea.

Example \(\PageIndex{1}\)

\[g_{n}=\left(\begin{array}{l}
g_{n}[1] \\
g_{n}[2]
\end{array}\right)=\left(\begin{array}{c}
1+\frac{1}{n} \\
2-\frac{1}{n}
\end{array}\right) \nonumber \]

First we find the following limits for our two \(g_n\)'s:

\[\begin{array}{l}
\operatorname{limit}_{n \rightarrow \infty} g_{n}[1]=1 \\
\operatorname{limit}_{n \rightarrow \infty} g_{n}[2]=2
\end{array} \nonumber \]

Therefore we have the following,

\[\operatorname{limit}_{n \rightarrow \infty} g_{n}=\boldsymbol{g} \nonumber \]

pointwise, where \(\boldsymbol{g}=\left(\begin{array}{l}
1 \\
2
\end{array}\right)\).

Example \(\PageIndex{2}\)

\[g_{n}(t)=\frac{t}{n}, t \in \mathbb{R} \nonumber \]

As done above, we first want to examine the limit

\[\operatorname{limit}_{n \rightarrow \infty} g_{n}\left(t_{0}\right)=\operatorname{limit}_{n \rightarrow \infty} \frac{t_{0}}{n}=0 \nonumber \]

where \(t_{0} \in \mathbb{R}\). Thus \(\operatorname{limit}_{n \rightarrow \infty} g_{n}=g\) pointwise where \(g(t)=0\) for all \(t \in \mathbb{R}\).

Norm Convergence

The sequence (Section 16.2) \(\left.\left\{g_{n}\right\}\right|_{n=1} ^{\infty}\) converges to \(\boldsymbol{g}\) in norm if \(\operatorname{limit}_{n \rightarrow \infty}\left\|g_{n}-g\right\|=0\). Here \(\|\cdot\|\) is the norm \(Section 15.3) of the corresponding vector space of \(g_n\)'s. Intuitively this means the distance between vectors \(g_n\) and \(\boldsymbol{g}\) decreases to \(0\).

Example \(\PageIndex{3}\)

\[g_{n}=\left(\begin{array}{c}
1+\frac{1}{n} \\
2-\frac{1}{n}
\end{array}\right) \nonumber \]

Let \(\boldsymbol{g}=\left(\begin{array}{l}
1 \\
2
\end{array}\right)\)

\[\begin{aligned}
\left\|g_{n}-\boldsymbol{g}\right\| &=\sqrt{\left(1+\frac{1}{n}-1\right)^{2}+\left(2-\frac{1}{n}\right)^{2}} \\
&=\sqrt{\frac{1}{n^{2}}+\frac{1}{n^{2}}} \\
&=\frac{\sqrt{2}}{n}
\end{aligned} \nonumber \]

Thus \(\operatorname{limit}_{n \rightarrow \infty}\left\|g_{n}-\boldsymbol{g}\right\|=0\). Therefore \(g_{n} \rightarrow \boldsymbol{g}\) in norm.

Example \(\PageIndex{4}\)

\[g_{n}(t)=\left\{\begin{array}{ll}
\frac{t}{n} & \text { if } 0 \leq t \leq 1 \\
0 & \text { otherwise }
\end{array}\right. \nonumber \]

Let \(g(t)=0\) for all \(t\).

\[\begin{aligned}
\left\|g_{n}(t)-g(t)\right\| &=\int_{0}^{1} \frac{t^{2}}{n^{2}} \mathrm{d} t \\
&=\left.\frac{t^{3}}{3 n^{2}}\right|_{n=0} ^{1} \\
&=\frac{1}{3 n^{2}}
\end{aligned} \nonumber \]

Thus \(\operatorname{limit}_{n \rightarrow \infty}\left\|g_{n}(t)-g(t)\right\|=0\) Therefore, \(g_{n}(t) \rightarrow g(t)\) in norm.

Pointwise vs. Norm Convergence

Theorem \(\PageIndex{1}\)

For \(\mathbb{R}^m\), pointwise and norm convergence are equivalent.

Proof: Pointwise ⇒ Norm

\[g_{n}[i] \rightarrow g[i] \nonumber \]

Assuming the above, then

\[\left(\left\|g_{n}-\boldsymbol{g}\right\|\right)^{2}=\sum_{i=1}^{m}\left(g_{n}[i]-g[i]\right)^{2} \nonumber \]

Thus,

\[\begin{aligned}
\operatorname{limit}_{n \rightarrow \infty}\left(\left\|g_{n}-\boldsymbol{g}\right\|\right)^{2} &=\operatorname{limit}_{n \rightarrow \infty} \sum_{i=1}^{m} 2 \\
&=\sum_{i=1}^{m} \operatorname{limit}_{n \rightarrow \infty} 2 \\
&=0
\end{aligned} \nonumber \]

Proof: Norm ⇒ Pointwise

\[\left\|g_{n}-\boldsymbol{g}\right\| \rightarrow 0 \nonumber \]

\begin{aligned}
\operatorname{limit}_{n \rightarrow \infty} \sum_{i=1}^{m} 2 &=\sum_{i=1}^{m} \operatorname{limit}_{n \rightarrow \infty} 2 \\
&=0
\end{aligned}

Since each term is greater than or equal zero, all '\(m\)' terms must be zero. Thus,

\[\operatorname{limit}_{n \rightarrow \infty} 2=0 \nonumber \]

for all \(i\). Therefore,

\[g_n \rightarrow \boldsymbol{g} \quad \text{ pointwise } \nonumber \]

Note

In infinite dimensional spaces the above theorem is no longer true. We prove this with counter examples shown below.

Counter Examples

Example \(\PageIndex{5}\): Pointwise \(\nRightarrow\) Norm

We are given the following function:

\[g_{n}(t)=\left\{\begin{array}{l}
n \text { if } 0<t<\frac{1}{n} \\
0 \text { otherwise }
\end{array}\right. \nonumber \]

Then \(\operatorname{limit}_{n \rightarrow \infty} g_{n}(t)=0\). This means that,

\[g_{n}(t) \rightarrow g(t) \nonumber \]

where for all \(t\) \(g(t)=0\).

Now,

\[\begin{aligned}
\left(\left\|g_{n}\right\|\right)^{2} &=\int_{-\infty}^{\infty}\left(\left|g_{n}(t)\right|\right)^{2} \mathrm{d} t \\
&=\int_{0}^{\frac{1}{n}} n^{2} \mathrm{d} t \\
&=n \rightarrow \infty
\end{aligned} \nonumber \]

Since the function norms blow up, they cannot converge to any function with finite norm.

Example \(\PageIndex{6}\): Norm \(\nRightarrow\) Pointwise

We are given the following function:

\[g_{n}(t)=\left\{\begin{array}{l}1 \text { if } 0<t<\frac{1}{n} \text { if } n \text { is even } \\ 0 \text { otherwise }\end{array}\right. \nonumber \]

\[g_{n}(t)=\left\{\begin{array}{l}-1 \text { if } 0<t<\frac{1}{n} \text { if } n \text { is odd } \\ 0 \text { otherwise }\end{array}\right. \nonumber \]

Then,

\[\left\|g_{n}-g\right\|=\int_{0}^{\frac{1}{n}} 1 \mathrm{d} t=\frac{1}{n} \rightarrow 0 \nonumber \]

where \(g(t)=0\) for all \(t\). Therefore,

\[g_n \rightarrow g \quad \text{in norm} \nonumber \]

However, at \(t=0\), \(g_n(t)\) oscillates between -1 and 1, and so it does not converge. Thus, \(g_n(t)\) does not converge pointwise.