Alejandro Atkins

2022-04-29

Let $f:\mathbb{R}\Rightarrow \mathbb{R}$ be a differentiable function, and suppose $f={f}^{\prime}$ and $f\left(0\right)=1$ . Then prove $f\left(x\right)\ne 0$ for all $x\in \mathbb{R}$

kubistiedt

Beginner2022-04-30Added 17 answers

Step 1

Similar to the work of yearning4pi on this topic:

With

1) ${f}^{\prime}\left(x\right)=f\left(x\right),\text{}f\left(0\right)=1,$

we consider the function ${e}^{-x}f\left(x\right)$; we have

2) ${\left({e}^{-x}f\left(x\right)\right)}^{\prime}=-{e}^{-x}f\left(x\right)+{e}^{-x}{f}^{\prime}\left(x\right)={e}^{-x}(-f\left(x\right)+{f}^{\prime}\left(x\right))=0$

in light of (1); thus

3) ${e}^{-x}f\left(x\right)=c,;\text{a constant},$

whence

4) $f\left(x\right)=c{e}^{x};$

then

5) $c=c{e}^{0}=f\left(0\right)=1,$

so that

6) $f\left(x\right)={e}^{x};$

clearly

7) ${e}^{x}=\sum _{0}^{\mathrm{\infty}}\frac{{x}^{n}}{n!}>0\text{}\text{for}\text{}x\ge 0,$

and if $x<0,\text{}-x0$, so

8) ${e}^{-x}>0;$

but

9) ${e}^{-x}{e}^{x}={e}^{-x+x}={e}^{0}=1\Rightarrow {e}^{x}=\frac{1}{{e}^{-x}}>0;$

thus

10) $\mathrm{\forall}x\in B\mathbf{R},;f\left(x\right)={e}^{x}>0.$

And then, of course, we always have Picard-Lindeloef as invoked by our colleague Severin Schraven.

Marely Anthony

Beginner2022-05-01Added 8 answers

Step 1

There is a shorter proof if you know the Picard-Lindelöf theorem. Assume that there exists$x}_{0}\in \mathbb{R$ such that $f\left({x}_{0}\right)=0$ . Then f solves the ODE

$\{\begin{array}{ll}{f}^{\prime}& =f\\ f({x}_{0})& =0.\end{array}$

Note that$g\equiv 0$ also solves the ODE above, but by the uniqueness part of Picard-Lindelöf we would have $f=g$ which contradicts $f\left(0\right)=1$ .

There is a shorter proof if you know the Picard-Lindelöf theorem. Assume that there exists

Note that

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