Difference between revisions of "Delta differentiable implies continuous"

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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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==Theorem==
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<strong>[[Delta derivative implies continuous|Theorem]]:</strong> If $f$ is $\Delta$-differentiable at $t$, then $f$ is [[continuity | continuous]] at $t$.
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If $f$ is [[Delta derivative|$\Delta$-differentiable]] at $t$, then $f$ is [[continuity | continuous]] at $t$.
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<div class="mw-collapsible-content">
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<strong>Proof:</strong>  █
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==Proof==
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</div>
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</div>
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Supposing \( f \) is differentiable, given that, by Definition 1.10, $ \epsilon^* > 0 $ there is a neighborhood \( U \) \( (U = (t - \delta, t + \delta) \cap \mathbb{T}, \text{ for some } \delta > 0) \) of \( t \) such that \( |t - s| < \delta \) and \( \left|f(\sigma(t)) - f(s) - f^\Delta(t)[\sigma(t) - s]\right| < \epsilon^* \left|\sigma(t) - s\right| \).
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Now, notice that
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$$ |f(t) - f(s)| = \left|f(t) - f(s) + f(\sigma(t)) - f(\sigma(t)) + f^\Delta(t)[\sigma(t) - s] - f^\Delta(t)[\sigma(t) - s]\right| $$
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Since \( \mu(t) = \sigma(t) - t \), then
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$$
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    \begin{array}{rcl}
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        |f(t) - f(s)| & < & \left|f(t) - f(s) + f(\sigma(t)) - f(\sigma(t)) + f^\Delta(t)[\sigma(t) - s] - f^\Delta(t)[\sigma(t) - s]\right| \\
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                      & = & |f(t) - f(s) + f(\sigma(t)) - f(\sigma(t)) + f^\Delta(t)[\sigma(t) + t - t - s] - f^\Delta(t)[\sigma(t) - s]| \\
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                      & = & |f(t) - f(s) + f(\sigma(t)) - f(\sigma(t)) + f^\Delta(t)[\mu(t) + t - s] - f^\Delta(t)[\sigma(t) - s]| \\
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                      & = & |f(t) - f(s) + f(\sigma(t)) - f(\sigma(t)) + f^\Delta(t)\mu(t) + f^\Delta(t)[t - s] - f^\Delta(t)[\sigma(t) - s]| \\
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                      & = & |f(\sigma(t)) - f(s) - f^\Delta(t)[\sigma(t) - s] - f(\sigma(t))  + f(t) + f^\Delta(t)\mu(t) + f^\Delta(t)[t - s]| \\
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                      & \leq & |f(\sigma(t)) - f(s) - f^\Delta(t)[\sigma(t) - s]| + |- f(\sigma(t)) + f(t) + f^\Delta(t)\mu(t)| + |f^\Delta(t)[t - s]| \\
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                      & = & |f(\sigma(t)) - f(s) - f^\Delta(t)[\sigma(t) - s]| + |f(\sigma(t)) - f(t) - f^\Delta(t)\mu(t)| + |f^\Delta(t)[t - s]| \\
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                      & \leq & \epsilon^*|\sigma(t) - s| + \epsilon^*|\mu(t)| + |f^\Delta(t)| \cdot |t -s|
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    \end{array}
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$$
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But, observe that
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$$
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    \begin{array}{rcl}
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        |\sigma(t) - s| & = & |\sigma(t) - s + t - t| \\
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                        & = & |\sigma(t) - t + t - s| \\
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                        & \leq & |\sigma(t) - t| + |t - s| \\
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                        & = & |\mu(t)| + |t - s| \\
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    \end{array}
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$$
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Thus, we have
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$$
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    \begin{array}{rcl}
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        |f(t) - f(s)| & \leq & \epsilon^*|\sigma(t) - s| + \epsilon^*|\mu(t)| + |f^\Delta(t)| \cdot |t -s| \\
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                      & \leq & \epsilon^*(|\mu(t)| + |t - s|) + \epsilon^*|\mu(t)| + |f^\Delta(t)| \cdot |t - s|
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    \end{array}
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$$
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However, taking \( \delta = \epsilon^* \) and knowing that \( |t - s| < \delta \) (by hypothesis), we have
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$$
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    \begin{array}{rcl}
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        |f(t) - f(s)| & \leq & \epsilon^*(|\mu(t)| + |t - s|) + \epsilon^*|\mu(t)| + |f^\Delta(t)| \cdot |t - s| \\
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                      & < & \epsilon^*(|\mu(t)| + |t - s|) + \epsilon^*|\mu(t)| + |f^\Delta(t)| \epsilon^* \\
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                      & = & \epsilon^*|\mu(t)| + \epsilon^*|t - s| + \epsilon^*|\mu(t)| + |f^\Delta(t)| \epsilon^* \\
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                      & = & \epsilon^*[|\mu(t)| + |t - s| + |\mu(t)| + |f^\Delta(t)|] \\
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                      & = & \epsilon^*[2|\mu(t)| + |t - s| + |f^\Delta(t)|]
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    \end{array}
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$$
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Furthermore, being aware that \( |\mu(t)| = \mu(t) \) (because \( \mu(t) \) is a function with non negative image) and considering (without loss of generality) \( \epsilon^* < 1 \), it follows that
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$$
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    \begin{array}{rcl}
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        |f(t) - f(s)| & < & \epsilon^*[2|\mu(t)| + |t - s| + |f^\Delta(t)|] \\
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                      & = & \epsilon^*[2\mu(t) + |t - s| + |f^\Delta(t)|] \\
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                      & < & \epsilon^*[2\mu(t) + 1 + |f^\Delta(t)|]
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    \end{array}
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$$
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Thus, since \( 2\mu(t) \geq 0 \), \( |t - s| \geq 0 \) and \( |f^\Delta(t)| \geq 0 \), we can let \( \epsilon = \epsilon^*[2\mu(t) + 1 + |f^\Delta(t)|] \). Hence, we get
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$$ |f(t) - f(s)| < \epsilon^*[2\mu(t) + 1 + |f^\Delta(t)|] = \epsilon $$
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Therefore, \( f \) is continuous at \( t \).
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==References==
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* {{BookReference|Dynamic Equations on Time Scales|2001|Martin Bohner|author2=Allan Peterson|prev=Delta derivative|next=Delta derivative at right-scattered}}: Theorem $1.16 (i)$
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[[Category:Theorem]]
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[[Category:Unproven]]

Latest revision as of 01:14, 15 March 2022

Theorem

If $f$ is $\Delta$-differentiable at $t$, then $f$ is continuous at $t$.

Proof

Supposing \( f \) is differentiable, given that, by Definition 1.10, $ \epsilon^* > 0 $ there is a neighborhood \( U \) \( (U = (t - \delta, t + \delta) \cap \mathbb{T}, \text{ for some } \delta > 0) \) of \( t \) such that \( |t - s| < \delta \) and \( \left|f(\sigma(t)) - f(s) - f^\Delta(t)[\sigma(t) - s]\right| < \epsilon^* \left|\sigma(t) - s\right| \).


Now, notice that

$$ |f(t) - f(s)| = \left|f(t) - f(s) + f(\sigma(t)) - f(\sigma(t)) + f^\Delta(t)[\sigma(t) - s] - f^\Delta(t)[\sigma(t) - s]\right| $$

Since \( \mu(t) = \sigma(t) - t \), then

$$ \begin{array}{rcl} |f(t) - f(s)| & < & \left|f(t) - f(s) + f(\sigma(t)) - f(\sigma(t)) + f^\Delta(t)[\sigma(t) - s] - f^\Delta(t)[\sigma(t) - s]\right| \\ & = & |f(t) - f(s) + f(\sigma(t)) - f(\sigma(t)) + f^\Delta(t)[\sigma(t) + t - t - s] - f^\Delta(t)[\sigma(t) - s]| \\ & = & |f(t) - f(s) + f(\sigma(t)) - f(\sigma(t)) + f^\Delta(t)[\mu(t) + t - s] - f^\Delta(t)[\sigma(t) - s]| \\ & = & |f(t) - f(s) + f(\sigma(t)) - f(\sigma(t)) + f^\Delta(t)\mu(t) + f^\Delta(t)[t - s] - f^\Delta(t)[\sigma(t) - s]| \\ & = & |f(\sigma(t)) - f(s) - f^\Delta(t)[\sigma(t) - s] - f(\sigma(t)) + f(t) + f^\Delta(t)\mu(t) + f^\Delta(t)[t - s]| \\ & \leq & |f(\sigma(t)) - f(s) - f^\Delta(t)[\sigma(t) - s]| + |- f(\sigma(t)) + f(t) + f^\Delta(t)\mu(t)| + |f^\Delta(t)[t - s]| \\ & = & |f(\sigma(t)) - f(s) - f^\Delta(t)[\sigma(t) - s]| + |f(\sigma(t)) - f(t) - f^\Delta(t)\mu(t)| + |f^\Delta(t)[t - s]| \\ & \leq & \epsilon^*|\sigma(t) - s| + \epsilon^*|\mu(t)| + |f^\Delta(t)| \cdot |t -s| \end{array} $$

But, observe that

$$ \begin{array}{rcl} |\sigma(t) - s| & = & |\sigma(t) - s + t - t| \\ & = & |\sigma(t) - t + t - s| \\ & \leq & |\sigma(t) - t| + |t - s| \\ & = & |\mu(t)| + |t - s| \\ \end{array} $$

Thus, we have

$$ \begin{array}{rcl} |f(t) - f(s)| & \leq & \epsilon^*|\sigma(t) - s| + \epsilon^*|\mu(t)| + |f^\Delta(t)| \cdot |t -s| \\ & \leq & \epsilon^*(|\mu(t)| + |t - s|) + \epsilon^*|\mu(t)| + |f^\Delta(t)| \cdot |t - s| \end{array} $$


However, taking \( \delta = \epsilon^* \) and knowing that \( |t - s| < \delta \) (by hypothesis), we have

$$ \begin{array}{rcl} |f(t) - f(s)| & \leq & \epsilon^*(|\mu(t)| + |t - s|) + \epsilon^*|\mu(t)| + |f^\Delta(t)| \cdot |t - s| \\ & < & \epsilon^*(|\mu(t)| + |t - s|) + \epsilon^*|\mu(t)| + |f^\Delta(t)| \epsilon^* \\ & = & \epsilon^*|\mu(t)| + \epsilon^*|t - s| + \epsilon^*|\mu(t)| + |f^\Delta(t)| \epsilon^* \\ & = & \epsilon^*[|\mu(t)| + |t - s| + |\mu(t)| + |f^\Delta(t)|] \\ & = & \epsilon^*[2|\mu(t)| + |t - s| + |f^\Delta(t)|] \end{array} $$


Furthermore, being aware that \( |\mu(t)| = \mu(t) \) (because \( \mu(t) \) is a function with non negative image) and considering (without loss of generality) \( \epsilon^* < 1 \), it follows that

$$ \begin{array}{rcl} |f(t) - f(s)| & < & \epsilon^*[2|\mu(t)| + |t - s| + |f^\Delta(t)|] \\ & = & \epsilon^*[2\mu(t) + |t - s| + |f^\Delta(t)|] \\ & < & \epsilon^*[2\mu(t) + 1 + |f^\Delta(t)|] \end{array} $$


Thus, since \( 2\mu(t) \geq 0 \), \( |t - s| \geq 0 \) and \( |f^\Delta(t)| \geq 0 \), we can let \( \epsilon = \epsilon^*[2\mu(t) + 1 + |f^\Delta(t)|] \). Hence, we get

$$ |f(t) - f(s)| < \epsilon^*[2\mu(t) + 1 + |f^\Delta(t)|] = \epsilon $$

Therefore, \( f \) is continuous at \( t \).

References

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