infinite vector space, derivative formula

Question

I'm presuming to make a major edit. The question as asked is badly garbled, and will likely be closed. That would be too bad, because the actual question in interesting and useful. Hence this revision.

Original, as incorrectly corrected:

How would you show that, given the infinite vector space entitled $V$ : $C^\infty ( \mathbb{R} , \mathbb{C} )$, and the endomorphism that associate to a function its derivative named D, that :

$$\forall f \in V : e^\lambda D^n ( e^{-\lambda} f ) = (D - \lambda Id_V ) ^n (f) $$

[Comment: It's not clear here whether $e^\lambda$ is supposed to be a constant, or we're regarding $\lambda$ as the variable and differentiating with respect to $\lambda$. But with either interpretation the assertion is simply false.]

I have no clue at all.

Thank you

Edit: The actual question is this:

For each scalar $\lambda$ define a function $e_\lambda$ by $e_\lambda(t)=e^{\lambda t}$. Show that $e_\lambda D^n(e_{-\lambda}f)=(D-\lambda I)^nf$.

(How do I know that's what the question was supposed to be? It's the same as the original original, except with a definition of $e_\lambda$ provided; the definition is perfectly natural, and makes the assertion both correct and interesting. Also the revised version is a standard and in some sense well known fact: For $n=1$ the identity is just an "abstract" or "operator-theoretic" formulation of the standard algorithm for solving the differential equation $y'-\lambda y=g$. If the idea that the OP would consistently write $e^\lambda$ in place of $e^{\lambda t}$ semms implausible I conjecture you've never taught differential equations...)

Answer 1

The original version of the question failed to define the notation $e_\lambda$. Curiously, the "correction" got the definition totally wrong.

The actual question is this:

For each scalar $\lambda$ define a function $e_\lambda$ by $e_\lambda(t)=e^{\lambda t}$. Show that $e_\lambda D^n(e_{-\lambda}f)=(D-\lambda I)^nf$.

In other words, if we let $m_\lambda$ denote the operation "multiply by $e_\lambda$", or $$(m_\lambda f)(t)=e^{\lambda t}f(t),$$the question is

Show that $(D-\lambda I)^n$ is "similar" to $D^n$, in particular $(D-\lambda I)^n=m_\lambda D^n m_\lambda^{-1}$.

For $n=1$ this is just the product rule:$$e^{\lambda t}\frac d{dt}(e^{-\lambda t}f(t))=e^{\lambda t}(e^{-\lambda t}f'(t)-\lambda e^{-\lambda t}f(t))=f'(t)-\lambda f(t).$$

Hence for any natural number $n$ we have $$(D-\lambda I)^n=(m_\lambda Dm_\lambda^{-1})^n=m_\lambda D^n m_\lambda^{-1}.$$