Original Problem

If $f$ is an entire function such that $$ f(z+1)-f(z)=f'(z) $$ for all $z$. Is there a non-trivial solution? ($f(z)=az+b$ is trivial)

And here is something uncertainty

If we use Fourier transform, how to define it to ensure any entire function has a FT?

Classical FT is defined by $$ \mathcal{F}[f] = F(\xi) = \int_{-\infty}^{+\infty}f(z)\mathrm{e}^{\mathrm{i} \xi z} \mathrm{d} z. $$ This only work for $f \in L^1(\mathbb{R})$. (If improved, it can work for $f \in L^2(\mathbb{R})$.)

Original Problem

If $f$ is an entire function such that $$ f(z+1)-f(z)=f'(z) $$ for all $z$. Is there a non-trivial solution? ($f(z)=az+b$ is trivial)

And here is something uncertainty

If we use Fourier transform, how to define it to ensure any entire function has a FT?

Classical FT is defined by $$ \mathcal{F}[f] = F(\xi) = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{+\infty}f(z)\mathrm{e}^{-\mathrm{i} \xi z} \mathrm{d} z. $$ This only work for $f \in L^1(\mathbb{R})$. (If improved, it can work for $f \in L^2(\mathbb{R})$.)

I know $\mathcal{F}[\mathrm{e}^{sz}] = \sqrt{2 \pi} \delta(\xi - \mathrm{i}s)$, but I'm not sure about a general definition.

If $f$ is an entire function such that $$ f(z+1)-f(z)=f'(z) $$ for all $z$. Is there a non-trivial solution? ($f(z)=az+b$ is trivial)

Original Problem

If $f$ is an entire function such that $$ f(z+1)-f(z)=f'(z) $$ for all $z$. Is there a non-trivial solution? ($f(z)=az+b$ is trivial)

And here is something uncertainty

If we use Fourier transform, how to define it to ensure any entire function has a FT?

Classical FT is defined by $$ \mathcal{F}[f] = F(\xi) = \int_{-\infty}^{+\infty}f(z)\mathrm{e}^{\mathrm{i} \xi z} \mathrm{d} z. $$ This only work for $f \in L^1(\mathbb{R})$. (If improved, it can work for $f \in L^2(\mathbb{R})$.)