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Prove a property of limits of vector valued functions

by RoRi

Prove that

    \[ \lim_{t \to p} F(t) = A \iff \lim_{t \to p} \lVert F(t) - A \rVert = 0. \]

Proof. Let

    \[ F(t) = (f_1(t), \ldots, f_n(t)), \qquad \text{and} \qquad A = (a_1, \ldots, a_n). \]

Then,

    \begin{align*} \lim_{t \to p} F(t) = A && \iff && \lim_{t \to p} (f_1(t), \ldots, f_n (t)) &= (a_1, \ldots, a_n) \\ && \iff && \left( \lim_{t \to p} f_1(t), \ldots, \lim_{t \to p} f_n (t) \right) &= (a_1, \ldots, a_n). \end{align*}

This holds if and only if

    \[ \lim_{t \to p} f_i (t) = a_i \qquad \text{for } i = 1, \ldots, n. \]

From the usual limit this is true if and only if

    \[ \lim_{t \to p} | f_i (t) - a_i | = 0 \qquad \text{for } i = 1, \ldots, n. \]

This is if and only if

    \begin{align*} &&\left( \lim_{t \to p} |f_1 (t) - a|, \ldots, \lim_{t \to p} |f_n (t) - a| \right) &= (0, \ldots, 0) \\ \iff && \lim_{t \to p} |F(t) - A| = 0. \qquad \blacksquare \end{align*}

One comment

  1. S says:

    The problem statement is for the norm of the vector, not its absolute value. However, it’s just another iff that cab be easily proven when going to the limit definition.

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