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I would appreciate if somebody could help me with the following problem:

Find $f(x)$, given that:

$f \colon \mathbb{R} \rightarrow \mathbb{R}$, $f$ is continuous at $x=0$, and $f(2x)=2f(x)+x$

I tried but couldn’t get it that way.

Community
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Young
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    What way did you try? – user141592 Mar 01 '15 at 15:14
  • I use; definition conti- function – Young Mar 01 '15 at 15:18
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    what is a conti- function ? – Tlön Uqbar Orbis Tertius Mar 01 '15 at 15:21
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    And what does "conti- on $x = 0$" mean? – Calum Gilhooley Mar 01 '15 at 15:23
  • $\lim_{x\to o}f(x)=f(0)$ then $f(x)$ conti- on $x=0$ – Young Mar 01 '15 at 15:24
  • Then you should say "$f$ is continuous at $0$", or something to that effect. These details matter! And the question can be edited. – Calum Gilhooley Mar 01 '15 at 15:26
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    Hint: let $f(x)=g(x)+x\log_2x$, then $f$ is solution iff $g$ is continuous at $0$ and, for every $x$, $$g(2x)=2g(x).$$ Can you finish? – Did Mar 01 '15 at 15:49
  • @Did. $g(2x) = f(2x) - 2x \log_2 (2x) = 2f(x) + x - 2x\log_2 (2x) = (f(x) - x \log_2 (x) ) + (f(x) - x \log_2 (x) ) + x = 2g(x) + x$ did I made a mistake ? – Tlön Uqbar Orbis Tertius Mar 01 '15 at 15:58
  • "did I made a mistake ?" Yes, you wrote $2x\log_2(2x)=2x\log_2(x)$ instead of $2x\log_2(2x)=2x\log_2(x)+$ $____$. Unrelated: The correct Ansatz rather seems to be $$f(x)=g(x)+\tfrac12x\log_2x.$$ – Did Mar 01 '15 at 16:44
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    I'm surprised that still no-one has pointed out that $f(x) = x\log_2x$ doesn't work (it leads to $f(2x) = 2f(x) + 2x$), and that it has been upvoted 3 times. I assumed at the time that Herbert Quain was pointing out the error, but the formatting of his comment was too garbled for me to be sure. This whole thread, including my own contributions - and all the many downvoted and deleted answers - has been quite messy! – Calum Gilhooley Mar 01 '15 at 18:01
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    @CalumGilhooley "I'm surprised that still no-one has pointed out that f(x)=xlog2x doesn't work" And one can be surprised that you are surprised since this HAS been mentioned, in my last comment (hint: search for "Unrelated"). – Did Mar 01 '15 at 18:25
  • @Did: Ah, sorry (again) - having given up on the parsing of Herbert Quain's comment, I didn't pursue the parsing of your comment on that comment (which has some box-like thing in it). But still, I share your surprise that I didn't notice the "unrelated" bit - was it in there from the start, or did you edit the comment? (If the latter, then I have some excuse for not noticing the unrelated addition.) What a mess it's all been! (Incidentally, my answer wasn't based on your comment - I just guessed at what $g(x)$ could be.) P.S. At least I see that the OP has now been edited - good. – Calum Gilhooley Mar 01 '15 at 19:20
  • @Young For your interest, an important part of the proof is to show that the solutions provided are the only ones. Just so that you will not be surprised by your TA's reaction if you hand in back the accepted answer to them... – Did Mar 02 '15 at 07:38

2 Answers2

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I have an ansatz, but I can neither show how to derive it from your functional equation nor prove that it is the only solution. Anyway, here it is:

$$f(x)=ax+\dfrac{x\ln(x)}{\ln{4}}$$

Which gives: \begin{align}f(2x)&=2ax+\dfrac{2x\ln{(2x)}}{\ln{4}}\\&=2ax+\dfrac{2x\ln{x}}{\ln{4}}+\dfrac{2x\ln{2}}{\ln{4}}\\&=2f(x)+2x\cdot\dfrac{1}{2}\\&=2f(x)+x\end{align}

as requested.

(Note that to move from line 1 to 2, we used $\ln{(ab)}=\ln{a}+\ln{b}$, and from line 2 to 3, $\ln{4}=\ln{2^2}=2\ln{2}$).

Demosthene
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Hint: put $g(x) = \frac{f(x)}{x}$ ($x \ne 0$), and define $f(0)$ suitably.

To be more explicit, consider the function $f$ defined by: $$ f(x) = \begin{cases} 0 & (x = 0) \\ \frac{x}{2}\log_2\left\lvert{x}\right\rvert & (x \ne 0) \end{cases} $$ Every function $f_a:x\mapsto f(x)+ax$ solves the problem and is continuous everywhere.

Did
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Calum Gilhooley
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  • $f(0) = 0$. Did you mean "define $g(0)$" suitably ? – Tlön Uqbar Orbis Tertius Mar 01 '15 at 16:15
  • @HerbertQuain: Sorry, I knew I'd put that clumsily, but I'm still not sure how to rewrite it! But I didn't mean "define $g(0)$"; $g(x)$ only needs to be defined for $x \ne 0$. Perhaps clearer is: "Determine $f(0)$, and for $x \ne 0$, write $g(x) = f(x)/x$." – Calum Gilhooley Mar 01 '15 at 16:18
  • Ok. I can't still figure how you solve the problem, I already tried this solution and it didn't work. Of course, if we suppose that $f$ is derivable at $0$ (or if we prove it), it will be clear that $f'(0) = f'(0) + 1/2$, leading to a contradiction. But I can't go further. – Tlön Uqbar Orbis Tertius Mar 01 '15 at 16:21
  • There is always the possibility that I've totally screwed up, of course! But it looks very simple and clear on paper (in the end). I'll triple-check ... One thing I notice is that I didn't even look at $f(x)$ for $x < 0$. It's certainly OK for $x \geqslant 0$. – Calum Gilhooley Mar 01 '15 at 16:23
  • ok, triple-check then write it ^^ – Tlön Uqbar Orbis Tertius Mar 01 '15 at 16:26
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    This cannot be the unique continuous solution since every $g(x)=ax$ is continuous and solves $g(2x)=2g(x)$. – Did Mar 01 '15 at 16:49
  • @Did: I posted an immediate reply to your comment, but it seems to have disappeared into the ether. (I'm not used to this website yet! I expect I did something silly.) Anyway, I then edited my answer, to correct the error you had pointed out. – Calum Gilhooley Mar 01 '15 at 17:12
  • I took upon me to revise your answer, keeping only the parts (that I saw as) interesting mathematically and adding the parts (that I saw as) missing mathematically. If the result does not suit you, just go back to a previous version. http://math.stackexchange.com/posts/1170478/revisions What is apparent after this revision is that, although being accepted, this answer provides some solutions but does not prove these are the only ones. And yet, I am ready to bet that, for the TA who asked this, this is the part they are waiting for... – Did Mar 02 '15 at 07:34
  • You probably did the right thing! Although I'm not exactly happy about it, there's probably no better solution to the mess that this little thread has got into - and at least it eases my embarrassment about my answer having been accepted, which it did not deserve to be! Anyone who, for whatever reason, wants to see what I actually wrote, and to distinguish your contribution, only needs to click on the "edited" link. The system works! I have had finer hours than this ... – Calum Gilhooley Mar 02 '15 at 08:33
  • What are you not happy with in my revision? (Unrelated: please use @.) – Did Mar 07 '15 at 11:34
  • @Did: Good Heavens, I thought this was over with! Far from objecting to your revision, I explicitly said you had "probably done the right thing", that there was "probably no better solution", that I was embarrassed, that my answer had not deserved to have been accepted, and that "the system works"! All this in 500 words, expressing to the best of my ability that I accepted the revision without demur. Evidently my ability in such social matters was inadequate, but then, we already knew that, because of the way I had allowed myself to be pressured into hasty posting, against my better judgement. – Calum Gilhooley Mar 07 '15 at 13:06
  • @Did: (Unrelated:) Sorry about the missing "@"! If it needs saying, I was still indescribably flustered by the whole episode, and I just forgot! Proof: I used "@" when replying to you previously. (Also unrelated:) By "500 words", of course I meant 500 characters. I'm becoming dreadfully flustered all over again. (Unrelated:) Anyone who is baffled by all this fuss over something so small can still read my original answer. There is nothing mathematical there wort – Calum Gilhooley Mar 07 '15 at 13:10
  • ... (Hit by the 5-minute editing limit! This is driving me insane, almost literally.) I was writing (when I accidentally hit the Enter key, in my distress) that there was nothing mathematical worth reading in my original answer, just some more detail about the horrible ridiculous embarrassing mess that this unimportant thread had got into, with much of the history of it having been deleted. It was the loss of all that information, and on top of it the disappearance from view of my information about that information, that I was "not exactly happy" about. But (until now) happily reconciled! – Calum Gilhooley Mar 07 '15 at 13:20
  • OK, OK, wow... As you say, everything is fine (and I was just asking by curiosity, nothing more). – Did Mar 07 '15 at 13:28
  • @Did: I'm relieved to hear that. I thought that by saying I was "not exactly happy", I must have inadvertently offended you; but instead, I had just failed to communicate my meaning clearly - which is par for the course, alas! No worries apart from that. – Calum Gilhooley Mar 07 '15 at 14:04