# Check whether $f \mapsto f+ \frac{df}{dx}$ is injective or surjective

Consider maps $$C^{\infty} \to C^{\infty}$$ s.t $$f \mapsto f+ \frac{df}{dx}$$. We have to check whether this map is injective or surjective.

My try: The map is clearly not injective as $$x$$ and $$x+e^{-x}$$ maps to $$x+1$$.

Now to check whether the map is surjective. Consider $$g \in C^{\infty}$$. Then I was thinking in this way that considering $$\int_0^xg$$ then $$f=g-\int_0^xg$$ now $$f+\frac{df}{dx}=g-\int_0^xg+\frac{dg}{dx}-g=-\int_0^xg+\frac{dg}{dx}$$ still I am not getting a proof whether it is surjective or not.

• More simply, $e^{-x}$ is mapped to $0$. – egreg May 18 '19 at 21:49

Checking surjectivity is the same as solving the ODE $$f'+f = g$$ for $$f$$ and seeing if you assume that $$g$$ is smooth, then $$f$$ is also smooth. This indeed happens, as we can solve the ODE by usual methods: since the solutions to $$f'+f=0$$ are of the form $$f(x) = Ce^{-x}$$, we try to look for general $$f$$ of the form $$f(x) = C(x)e^{-x}$$. Then $$g(x) = f'(x)+f(x) = C'(x)e^{-x}-C(x)e^{-x} + C(x)e^{-x} = C'(x)e^{-x}$$implies that $$C(x) = \int e^xf(x)\,{\rm d}x$$, and of course $$f(x) = e^{-x}\int_0^x e^tg(t)\,{\rm d}t$$is smooth if $$g$$ is.
• Could you change your Integration variable so it doesn't equal the Parameter of $x$? – SK19 May 18 '19 at 20:48
If you want $$f^\prime(x) + f(x) = g(x)$$, then multiplying by $$e^x$$ to get $$(e^x f(x))^\prime =e^xf^\prime(x) + e^x f(x) = e^x(f^\prime(x) + f(x)) = e^xg(x)$$ should let you find an $$f$$ that works.