# A functional analysis problem

1. May 30, 2012

### wdlang

see the attachment

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2. May 30, 2012

### HallsofIvy

Since, in the statement of the problem, f is only required to be continuous, not differentiable, what does f ' mean?

3. May 30, 2012

### wdlang

f is continuous means f is infinitely differentiable

4. May 30, 2012

### Whovian

No it doesn't. Assume $\displaystyle f\left(x\right)=\left|x\right|$. It's obviously continuous ($\displaystyle\forall a\in\mathbb{R}\ \left(\lim_{x\to a}\left|x\right|=\left|a\right|\right)$), but its derivative at 0 doesn't exist.

5. May 30, 2012

### wdlang

ok, i mean smooth

sorry for that

6. May 30, 2012

### Whovian

Okay. $\displaystyle\dfrac{x\cdot\left|x\right|}{2}$. It's smooth, but its derivative is |x|.

Though I'm pretty sure you mean continuously differentiable, which is ideally what we'd assume.

7. May 30, 2012

### brydustin

Actually, I think he DOES mean smooth, a function is smooth (by definition) if it is infinitely differentiable. So your "counter-example" is not valid; in other words, $\displaystyle\dfrac{x\cdot\left|x\right|}{2}$ is not smooth.

8. May 30, 2012

### HallsofIvy

"smooth" is not a standardized term. In some texts it means "infinitely differentiable" but in others it only means "the first derivative is continuous" and "infinitely smooth" is used to mean "infinitely differentiable". In some texts you will even see "sufficiently smooth" meaning "differentiable to whatever extent is necessary to prove this".

9. May 30, 2012

### pwsnafu

I've also seen "smooth" to mean "analytic" in the sense of Taylor series convergence, which would be separate from "infinitely differentiable".

10. May 30, 2012

### brydustin

OKAY! We get it, there are a lot of different terms for the same thing, and sometimes the same term means different things. Let's focus on the post's question, not some terms; all we need to know is that the function "f is continuous means f is infinitely differentiable", and that's all we need to know to attempt to find a solution. Focus on that.

11. May 30, 2012

### brydustin

Why take the absolute value squared, if the function is real valued? Just take the square of the function, right? Or is that the supremum norm of the function? i.e. $|f| = sup(f)$

12. May 31, 2012

### algebrat

It loosk like you might be able to make one of those proofs like considering the sequence of regions where |f|<1/n, then blah blah. Not sure if that'll work, but t's first thing that popped into my head.