In lectures we defined the fourier and fourier cosine (for even functions) transforms as follows:
$$ \widehat{f}(\omega)=\int_{-\infty}^{\infty} f(x) e^{-i \omega x} d x $$
$$ \widehat{f}_{c}(\omega)=\int_{0}^{\infty} f(x) \cos (\omega x) d x $$
It is clear that $$ \widehat{f}(\omega) = 2 \widehat{f}_{c}(\omega) $$
Now we also derived two formulas for taking the fourier transforms and fourier cos transforms of second derivatives:
$$ \mathcal{F}\left\{d^{n} f / d x^{n}\right\}=(i \omega)^{n} \widehat{f}(\omega) $$
$$ \mathcal{F}_{c}\left\{f^{\prime \prime}(x)\right\}=-f^{\prime}(0)-\omega^{2} \widehat{f}_{c}(\omega) $$
But equating them with taking into account the factor of 2 leads to:
$$ \mathcal{F}_{c}\left\{f^{\prime \prime}(x)\right\} = -f^{\prime}(0)-\omega^{2} \widehat{f}_{c}(\omega) = \frac{1}{2} \mathcal{F}\left\{f^{\prime \prime}(x)\right\} = -\omega^{2} \widehat{f}_{c}(\omega) $$
Which suggest that $f^{\prime}(0) = 0$, which is not necessarily true, am I assuming something incorrect here?
Best Answer
Consider an even function $f$, i.e. $f(-x) = f(x)$,
$$\begin{align*} f'(x) &= \frac{df(x)}{dx}\\ &= \frac{df(-x)}{dx}\\ &= \frac{df(-x)}{d(-x)}\frac{d(-x)}{dx}\\ &= -f'(-x) \end{align*}$$
This shows $f'$ is odd, and by substituting $x=0$, $$f'(0) = 0$$