I'm having trouble formulating a proof of the following theorem:
Let $f$ be a real-valued function having a continuous $(n - 1)$st derivative $f^{(n - 1)}$ for $t \geq 0$; and assume that $f, f^\prime, f^{\prime\prime}, \dots, f^{(n - 1)}$ are all of exponential order $e^{\alpha t}$. Suppose $f^{(n)}$ is piecewise continuous in every finite closed interval $0 \leq t \leq b$. Then $\mathcal{L} \{ f^{(n)} \}$ exists for $s > \alpha$ and$$\mathcal{L} \{f^{(n)}\} = s^n\mathcal{L} \{ f \} - s^{n - 1}f(0)-s^{n - 2}f^\prime(0) - s^{n-3} f^{\prime\prime}(0) - \ldots - f^{(n-1)}(0).$$
I think this would be an inductive reasoning proof, but I can't seem to prove the generalization.
I know that$$\mathcal{L} \{ f^\prime \} = s\mathcal{L} \{ f \} - f(0),$$and that$$\mathcal{L} \{ f^{(2)} \} = s^2\mathcal{L} \{ f \} - sf(0)-f^\prime(0)$$But how should I attempt to generalize this?
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$\begingroup$You've got the basis case $n=1$. If $g=f^{(k)}$ and $$\mathcal{L}(g) = \mathcal{L}(f^{(k)}) = s^k \mathcal{L}(f) - s^{k-1}f(0) - \dotsb - f^{(k-1)}(0), $$ then using the $n=1$ result gives \begin{align} \mathcal{L}(f^{(k+1)})=\mathcal{L}(g') &= s\mathcal{L}(g) - g(0) \\ &= s(s^k \mathcal{L}(f) - s^{k-1}f(0) - \dotsb - f^{(k-1)}(0)) - f^{(k)}(0) \\ &= s^{k+1} \mathcal{L}(f) - s^{k}f(0) - \dotsb - sf^{(k-1)}(0)) - f^{(k)}(0), \end{align} which is of the same form with $k \mapsto k+1$, so combining this with the basis case, induction implies the result holds for all positive integers $n$.
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