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extentending a differentiable function

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extentending a differentiable function

hi
suppose f:[a,b]->R is differentiable with sided derivatives at endpoints.
can we extend f to a differentiable function on R.
that is, is there any differentiable functions F:R->R so that
(forall x in [a,b]) F(x)=f(x)
?


thanks

Cf. the begin of the Taylor expansion f(x) = f(b)+f'(b)(x-b)+...

You are looking at it from the wrong direction. :)

If f:R->R is differentiable, then obviously so is the restriction of f to [a,b]. So this is your example.

But we are asking the reverse question. If f:[a,b]->R is differentiable, then can f be extended to entire real line? The answer is yes (shown by Pahio).

To show you that this is not trivial question, consider the same question, but f:(a,b)->R is defined on the *OPEN* interval. For example, let f:(0,1)->R be define by f(x)=1/x. This is well defined differential function which *CANNOT* be extended to entire line, because there is no (finite) limit of f(x) when x tends to 0. In other words "the extension problem" has negative solution for open intervals.

Many similar questions may be asked. For example, can every smooth function f:[a,b]->R be extended to a smooth function F:R->R? The answer is again YES, but this is a bit more complex. :)

I hope I cleared "the fog" a little bit. :) If not, then forget a said anything. :P

joking

you cleared the fog but without proof!

Yes. You may take the missing parts of the graph of F half-lines. E.g.
F(x) := f'(b)(x-b)+f(b) when x > b.

I hklass, I cannot understand the statement of your problem. You claim f(x)=F(x)fxFxf(x)=F(x) forall xxx in [a,b]ab[a,b]. Well, it is legitimate if, e.g. I consider f(x)=n=0ancosnx+bnsinnxfxsuperscriptsubscriptn0subscriptannxsubscriptbnnxf(x)=\sum_{{n=0}}^{\infty}a_{n}\cos nx+b_{n}\sin nx. May I to extend this function over all \mathbb{R}? YES! perucho

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