Example:A smooth function that is not real analytic
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Example
Let [ilmath]f:\mathbb{R}\rightarrow\mathbb{R} [/ilmath] be defined as follows:
- [math]f:x\mapsto\left\{\begin{array}{lr}e^{-\frac{1}{x} } & \text{if }x>0\\ 0 & \text{otherwise}\end{array}\right.[/math]
We will show this function is not real-analytic at [ilmath]0\in\mathbb{R} [/ilmath] but is smooth.
Proof
For [ilmath]x<0[/ilmath], [ilmath]f[/ilmath] is smooth
This is trivial, as [ilmath]f\vert_{\mathbb{R}_{<0} }\ident 0[/ilmath]
- TODO: The derivative of a constant is 0, the derivative of that is 0, and so forth - link to corresponding pages
For [ilmath]x>0[/ilmath], [ilmath]f[/ilmath] is smooth
We will show a slightly different result.
- Let [ilmath]P_m(y)[/ilmath] be a polynomial of order [ilmath]m[/ilmath] in the invariant [ilmath]y[/ilmath]. [ilmath]p_m:\mathbb{R}\rightarrow\mathbb{R} [/ilmath] is the map with [ilmath]p_m:y\mapsto p(y)[/ilmath] meaning evaluation at [ilmath]y\in\mathbb{R} [/ilmath].
We claim that:
- [math]\frac{\mathrm{d}^kf}{\mathrm{d}x^k}\eq p_{2k}(\tfrac{1}{x})\mathrm{e}^{-\frac{1}{x} } [/math]
Proof
To avoid any potential ambiguities Caveat:I'm not sure whether or not there'd be problems with a 0 degree polynomial, but I want to sidestep it we shall prove this by induction starting from [ilmath]k\eq 1[/ilmath].
The proof shall proceed as follows:
- Show that [math]\frac{df}{dx}\eq p_2(\tfrac{1}{x})e^{-\frac{1}{x} } [/math]
- Assume that [math]\frac{d^kf}{dx^k}\eq p_{2k}(\tfrac{1}{x})e^{-\frac{1}{x} } [/math] holds
- Show that [math]\left(\frac{d^kf}{dx^k}\eq p_{2k}(\tfrac{1}{x})e^{-\frac{1}{x} }\right)\implies\left(\frac{d^{k+1}f}{dx^{k+1} }\eq p_{2(k+1)}(\tfrac{1}{x})e^{-\frac{1}{x} }\right)[/math]
- I.E. show that the RHS of this is true (as by the nature of logical implication if the LHS is true - which we assume it is - then for it to imply the RHS the RHS must also be true.
Proof:
- Remember that here we are explicitly dealing with the [ilmath]x\in\mathbb{R}_{>0} [/ilmath] case. We are basically considering [ilmath]f:\mathbb{R}_{>0}\rightarrow\mathbb{R} [/ilmath] by [ilmath]f:x\mapsto e^{-\frac{1}{x} } [/ilmath]
- Show that [ilmath]\frac{df}{dx}\eq p_2(\frac{1}{x})e^{-\frac{1}{x} } [/ilmath] Caveat:I'm experimenting with differentiation notation here[Note 1]
- [math]\frac{df}{dx}\Big\vert_x\eq\frac{d}{dz}e^{z}\Big\vert_{z\eq -\frac{1}{x} }\cdot\frac{d}{dx}-(x^{-1})\Big\vert_x[/math][math]\eq e^{-\frac{1}{x} }\cdot(-1)\cdot\frac{d}{dx}x^{-1}\Big\vert_x[/math][math]\eq x^{-2}e^{-\frac{1}{x} } [/math]
- We see [math]\frac{df}{dx}\eq \left(\tfrac{1}{x}\right)^2e^{-\frac{1}{x} } [/math] or [math]\frac{df}{dx}\eq p_2(\tfrac{1}{x})e^{-\frac{1}{x} } [/math]
- Now we assume [ilmath]\dfrac{d^kf}{dx^k}\eq p_{2k}(\tfrac{1}{x})e^{-\frac{1}{x} } [/ilmath]
- We attempt to show that [ilmath]\left(\dfrac{d^kf}{dx^k}\eq p_{2k}(\tfrac{1}{x})e^{-\frac{1}{x} }\right)\implies\left(\dfrac{d^{k+1}f}{dx^{k+1} }\eq p_{2(k+1)}(\tfrac{1}{x})e^{-\frac{1}{x} } \right)[/ilmath]
- [math]\frac{d^{k+1}f}{dx^{k+1} }\eq\frac{d}{dx}\left[\frac{d^kf}{dx^k}\right]_x\eq\frac{d}{dx}\left[p_{2k}(\tfrac{1}{x})e^{-\frac{1}{x} }\right][/math][math]\eq e^{-\frac{1}{x} }\left(\frac{d}{dy}\Big[p_{2k}(y)\Big]_{y\eq-\frac{1}{x} }\cdot\frac{d}{dx}\Big[-x^{-1}\Big]\right)+p_{2k}(\tfrac{1}{x})\cdot\frac{d}{dx}\Big[e^{-\frac{1}{x} }\Big] [/math]
- Well if we differentiate a polynomial of order [ilmath]2k[/ilmath] we get a polynomial of order [ilmath]2k-1[/ilmath], so: Switching from [ilmath]p[/ilmath] to [ilmath]P[/ilmath] for the polynomial to make the subscripts more obvious
- [math]\frac{d^{k+1}f}{dx^{k+1} }\eq e^{-\frac{1}{x} }P_{2k-1}(\tfrac{1}{x})\cdot x^{-2} + P_{2k}(\tfrac{1}{x})\cdot x^{-2}e^{-\frac{1}{x} } [/math] [math]\eq e^{-\frac{1}{x} }\left(P_{2k-1}(\tfrac{1}{x})\cdot x^{-2}+P_{2k}(\tfrac{1}{x})\cdot x^{-2}\right)[/math]
- As [ilmath]x^{-1}\eq(\tfrac{1}{x})^2[/ilmath] we see:
- [math]\frac{d^{k+1}f}{dx^{k+1} }\eq e^{-\frac{1}{x} } \left(P_{2k+1}(\tfrac{1}{x}) + P_{2(k+1)}(\tfrac{1}{x}) \right)[/math] (note that [ilmath]2k+2\eq 2(k+1)[/ilmath] in the subscript of [ilmath]P[/ilmath])
- Thus: [math]\frac{d^{k+1}f}{dx^{k+1} }\eq P_{2(k+1)}(\tfrac{1}{x})e^{-\frac{1}{x} } [/math] - as expected.
- We have shown the induction hypothesis.
- [math]\frac{d^{k+1}f}{dx^{k+1} }\eq\frac{d}{dx}\left[\frac{d^kf}{dx^k}\right]_x\eq\frac{d}{dx}\left[p_{2k}(\tfrac{1}{x})e^{-\frac{1}{x} }\right][/math][math]\eq e^{-\frac{1}{x} }\left(\frac{d}{dy}\Big[p_{2k}(y)\Big]_{y\eq-\frac{1}{x} }\cdot\frac{d}{dx}\Big[-x^{-1}\Big]\right)+p_{2k}(\tfrac{1}{x})\cdot\frac{d}{dx}\Big[e^{-\frac{1}{x} }\Big] [/math]
Grade: A*
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Important work for manifolds, will probably crop up again!
Notes
- ↑ The details are as follows:
- [math]\frac{df}{dx} [/math] is itself a function that takes [ilmath]x'[/ilmath] to [math]\frac{df}{dx}\Big\vert_{x\eq x'} [/math]
- [math]\frac{df}{dx}\Big\vert_x[/math] is another way of writing [math]\frac{df}{dx} [/math] - anywhere where the variable we are differentiating with respect to is where we are differentiating at invokes this, and is actually a function in the "at" part.
- [math]\frac{df}{dx}\Big\vert_{x\eq z} [/math] is an expression for the derivative of [ilmath]f[/ilmath] (wrt [ilmath]x[/ilmath]) - at [ilmath]z[/ilmath] - note that it is an expression - not a constant:
- For example [math]\frac{d}{dx}x^3\Big\vert_{x\eq t}\eq 3t^2[/math] so [math]\frac{d}{dt}\frac{d}{dx}x^3\Big\vert_{x\eq t}\Big\vert_{t}\eq 6t[/math]
- [math]\frac{d}{dt}\Big[\text{whatever}\Big][/math] to mean [math]\frac{d}{dt}\text{whatever}\Big\vert_t[/math]
- [math]\frac{d}{dt}\Big[\text{whatever}\Big]_{t} [/math] to mean [math]\frac{d}{dt}\text{whatever}\Big\vert_t[/math], and
- [math]\frac{d}{dt}\Big[\text{whatever}\Big]_{t\eq p} [/math] to mean [math]\frac{d}{dt}\text{whatever}\Big\vert_{t\eq p} [/math]
- [math]\frac{d}{dt}\Big[\text{whatever}\Big]\Big\vert_{t\eq p} [/math] for example
