given that $V$ is an open subset of $\mathbb{C}$ and $z \in V$, calculate $\lim_{h\rightarrow 0} \dfrac {e^{f(z+h)}-e^{f(z)}}{f(z+h)- f(z)}$, if $f$ is known to be a continuous complex function in $V$. I know that the result is supposed to be $e^{f(z)}$, but I can't figure out the details in a rigorous way. Any help appreciated and many thanks in advance.
Asked
Active
Viewed 100 times
4
-
1What happens if you multiply by $\frac hh$? Hmm, maybe not, I guess you're not given that $f$ is differentiable... – abiessu Dec 05 '13 at 15:27
-
@aviessu It does not quite help, since $f$ is not assumed to be differentiable. – Andrés E. Caicedo Dec 05 '13 at 15:31
-
Set $w_0 = f(z)$ and $w = f(z+h)$. Note that the continuity of $f$ says $(h \to 0) \Rightarrow (w \to w_0)$. – Daniel Fischer Dec 05 '13 at 15:38
-
@DanielFischer right. Can you add it as an answer? – leo Dec 05 '13 at 15:41
-
@leo No, $f$ is not locally differentiable. The standard Cauchy's integral formula does not apply, $f$ is only assumed continuous, not holomorphic. – Andrés E. Caicedo Dec 05 '13 at 16:09
3 Answers
2
Hint: Write $w_0 = f(z)$ and $w = f(z+h)$. Note that the continuity of $f$ implies
$$(h\to 0) \Rightarrow (w \to w_0).$$
The continuity of $f$ in $z$ means that for any arbitrarily prescribed $\delta > 0$, we can find an $\eta > 0$ such that we have $\lvert f(z+h) - f(z)\rvert < \delta$ for all $h$ with $\lvert h\rvert < \eta$.
We have - if $f(z+h) \neq f(z)$ -
$$ \left\lvert \frac{e^{f(z+h)}- e^{f(z)}}{f(z+h)-f(z)} - e^{f(z)}\right\rvert = \left\lvert e^{f(z)}\right\rvert\cdot\left\lvert \frac{e^{f(z+h)-f(z)}-1}{f(z+h)-f(z)}-1\right\rvert. $$
Let us for brevity write $y = f(z+h)-f(z)$.
For $\lvert y\rvert < \frac12$, we have
$$\lvert e^y - 1-y\rvert = \left\lvert \sum_{n=2}^\infty \frac{y^n}{n!}\right\rvert \leqslant \sum_{n=2}^\infty \frac{\lvert y\rvert^n}{n!} \leqslant \frac{\lvert y\rvert^2}{2}\sum_{k=0}^\infty \lvert y\rvert^k = \frac{\lvert y\rvert^2}{2(1-\lvert y\rvert)} < \lvert y\rvert^2,$$
and therefore
$$ \left\lvert \frac{e^y-1}{y} - 1\right\rvert = \left\lvert \frac{e^y-1-y}{y}\right\rvert \leqslant \lvert y\rvert. $$
Now, given an arbitrary $\varepsilon > 0$, we set $\delta = \min \left\lbrace \frac12, \frac{\varepsilon}{\lvert e^{f(z)}\rvert}, \varepsilon\right\rbrace$, and for that $\delta$ choose $\eta > 0$ such that $\lvert f(z+h)-f(z)\rvert < \delta$ for $\lvert h\rvert < \eta$. Then we have
$$\left\lvert \frac{e^{f(z+h)}- e^{f(z)}}{f(z+h)-f(z)} - e^{f(z)}\right\rvert \leqslant \lvert e^{f(z)}\rvert \cdot \lvert f(z+h) - f(z)\rvert < \lvert e^{f(z)}\rvert\cdot\delta < \varepsilon$$
for $\lvert h\rvert < \eta$, so
$$\lim_{h\to 0} \frac{e^{f(z+h)}-e^{f(z)}}{f(z+h)-f(z)} = e^{f(z)},$$
subject to the restriction that we must have $f(z+h) \neq f(z)$.
Daniel Fischer
- 206,697
-
That is what I meant with 'I know the answer but can't prove it in a rigorous way'. To me, this needs refining, but I don't know how to exactly go about it... – user112105 Dec 05 '13 at 20:09
-
-
-
Daniel, rather than using the full power series, I think it may be simpler to argue with the error estimate for the first Taylor polynomial. This gives that there is a constant $C>0$ such that for $|y|$ small enough, $|e^y-1-y|\le C|y|^2$, which suffices. – Andrés E. Caicedo Dec 05 '13 at 21:33
1
HINT. Expand both numerator and denominator as Taylor series around h=0. Limit the expansion to first order, it will be sufficient. Can you continue from here ? All of this supposes that f(z) is differentiable.
Claude Leibovici
- 260,315
-
1If $f$ is differentiable, then just use the definition of the derivative and $(e^f)' = f' e^f$... – Najib Idrissi Dec 05 '13 at 20:53
0
This is much easier than it seems. Don't be thrown off by the usage of $f(z + h) - f(z)$, which hints at dividing by $h$. The only thing that is important is that $f(z + h) \to f(z)$ when $h \to 0$ and the definition of the derivative of the exponential function.
Concretely, use the following general fact, which follows straightforwardly from the $\epsilon$-$\delta$-definition of limits.
Theorem. If $\lim_{x \to x_0} a(x) = l$ and $\lim_{h \to h_0} b(h) = x_0$, then $\lim_{h \to h_0} a(b(h)) = l$.
Now apply this to $a(x) := \frac{\exp(x) - \exp(f(z))}{x - f(z)}$ with $x \to f(z)$ and $b(h) := f(z + h)$ with $h \to 0$. Note that $\lim_{h \to 0} b(h) = f(z)$ because $f$ is continuous and that $\lim_{x \to f(z)} a(x) = \exp'(f(z))$ by definition of $\exp'$, so $\lim_{x \to f(z)} a(x) = \exp(f(z))$. Consequently, using the theorem above, $\lim_{h \to 0} \frac{\exp(f(z + h)) - \exp(f(z))}{f(z+h) - f(z)} = \lim_{h \to 0} a(b(h)) = \exp(f(z))$.
Magdiragdag
- 15,049