Delta differentiable implies continuous

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 $1.16 (i)$