Difference between revisions of "Delta differentiable implies continuous"
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| − | + | ==Theorem== | |
| − | + | If $f$ is [[Delta derivative|$\Delta$-differentiable]] at $t$, then $f$ is [[continuity | 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== | ||
| + | * {{BookReference|Dynamic Equations on Time Scales|2001|Martin Bohner|author2=Allan Peterson|prev=Delta derivative|next=Delta derivative at right-scattered}}: Theorem $1.16 (i)$ | ||
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[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
- Martin Bohner and Allan Peterson: Dynamic Equations on Time Scales (2001)... (previous)... (next): Theorem $1.16 (i)$
