edited by
164 views
0 votes
0 votes

Let $C^{\infty }\left ( 0,1 \right )$ stand for the set of all real-valued functions on $\left ( 0,1 \right )$ that have derivatives of all orders. Then the map $C^{\infty }\left ( 0,1 \right )\rightarrow C^{\infty }\left ( 0,1 \right )$ given by

$$f \mapsto f+\frac{df}{dx}$$

is

  1. injective but not surjective
  2. surjective but not injective
  3. neither injective nor surjective
  4. both injective and surjective

 

edited by

Please log in or register to answer this question.

Answer:

Related questions

0 votes
0 votes
0 answers
1
soujanyareddy13 asked Aug 29, 2020
107 views
True/False Question :There exists a uniformly continuous function $f:\left ( 0,\infty \right )\rightarrow \left ( 0,\infty \right )$ such that $$\sum_{n=1 }^{\infty }\fra...
0 votes
0 votes
0 answers
2
soujanyareddy13 asked Aug 29, 2020
175 views
The following sum of numbers (expressed in decimal notation)$$1+11+111+\cdots +\underset{n}{\underbrace{11\dots1}}$$is equal to$\left ( 10^{n+1}-10-9n \right )/81$$\left ...
0 votes
0 votes
0 answers
3
soujanyareddy13 asked Aug 29, 2020
171 views
For $n\geq 1$, the sequence $\left \{ x_{n} \right \}^{\infty }_{n=1},$ where:$$x_{n}=1+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{n}}-2\sqrt{n}$$isdecreasingincreasingconst...