# 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:

• $f:x\mapsto\left\{\begin{array}{lr}e^{-\frac{1}{x} } & \text{if }x>0\\ 0 & \text{otherwise}\end{array}\right.$

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:

• $\frac{\mathrm{d}^kf}{\mathrm{d}x^k}\eq p_{2k}(\tfrac{1}{x})\mathrm{e}^{-\frac{1}{x} }$

#### 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:

1. Show that $\frac{df}{dx}\eq p_2(\tfrac{1}{x})e^{-\frac{1}{x} }$
2. Assume that $\frac{d^kf}{dx^k}\eq p_{2k}(\tfrac{1}{x})e^{-\frac{1}{x} }$ holds
3. Show that $\left(\frac{d^kf}{dx^k}\eq p_{2k}(\tfrac{1}{x})e^{-\frac{1}{x} }\right)\implies\left(\frac{d^{k+1}f}{dx^{k+1} }\eq p_{2(k+1)}(\tfrac{1}{x})e^{-\frac{1}{x} }\right)$
• 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]
1. 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]
• $\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$$\eq e^{-\frac{1}{x} }\cdot(-1)\cdot\frac{d}{dx}x^{-1}\Big\vert_x$$\eq x^{-2}e^{-\frac{1}{x} }$
• We see $\frac{df}{dx}\eq \left(\tfrac{1}{x}\right)^2e^{-\frac{1}{x} }$ or $\frac{df}{dx}\eq p_2(\tfrac{1}{x})e^{-\frac{1}{x} }$
2. Now we assume [ilmath]\dfrac{d^kf}{dx^k}\eq p_{2k}(\tfrac{1}{x})e^{-\frac{1}{x} } [/ilmath]
3. 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]
• $\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]$$\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]$
• 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
• $\frac{d^{k+1}f}{dx^{k+1} }\eq e^{-\frac{1}{x} }P_{2k-1}(\tfrac{1}{x})\cdot x^{-2} + P_{2k}(\tfrac{1}{x})\cdot x^{-2}e^{-\frac{1}{x} }$ $\eq e^{-\frac{1}{x} }\left(P_{2k-1}(\tfrac{1}{x})\cdot x^{-2}+P_{2k}(\tfrac{1}{x})\cdot x^{-2}\right)$
• As [ilmath]x^{-1}\eq(\tfrac{1}{x})^2[/ilmath] we see:
• $\frac{d^{k+1}f}{dx^{k+1} }\eq e^{-\frac{1}{x} } \left(P_{2k+1}(\tfrac{1}{x}) + P_{2(k+1)}(\tfrac{1}{x}) \right)$ (note that [ilmath]2k+2\eq 2(k+1)[/ilmath] in the subscript of [ilmath]P[/ilmath])
• Thus: $\frac{d^{k+1}f}{dx^{k+1} }\eq P_{2(k+1)}(\tfrac{1}{x})e^{-\frac{1}{x} }$ - as expected.
• We have shown the induction hypothesis.
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Important work for manifolds, will probably crop up again!

## Notes

1. The details are as follows:
• $\frac{df}{dx}$ is itself a function that takes [ilmath]x'[/ilmath] to $\frac{df}{dx}\Big\vert_{x\eq x'}$
• $\frac{df}{dx}\Big\vert_x$ is another way of writing $\frac{df}{dx}$ - 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.
• $\frac{df}{dx}\Big\vert_{x\eq z}$ 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 $\frac{d}{dx}x^3\Big\vert_{x\eq t}\eq 3t^2$ so $\frac{d}{dt}\frac{d}{dx}x^3\Big\vert_{x\eq t}\Big\vert_{t}\eq 6t$
We may also use:
• $\frac{d}{dt}\Big[\text{whatever}\Big]$ to mean $\frac{d}{dt}\text{whatever}\Big\vert_t$
• $\frac{d}{dt}\Big[\text{whatever}\Big]_{t}$ to mean $\frac{d}{dt}\text{whatever}\Big\vert_t$, and
• $\frac{d}{dt}\Big[\text{whatever}\Big]_{t\eq p}$ to mean $\frac{d}{dt}\text{whatever}\Big\vert_{t\eq p}$
Rather than:
• $\frac{d}{dt}\Big[\text{whatever}\Big]\Big\vert_{t\eq p}$ for example