Difference between revisions of "Example:A smooth function that is not real analytic"

Latest revision as of 04:25, 27 November 2016

Example

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ 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 $0\in\mathbb{R}$ but is smooth.

Proof

For $x<0$, $f$ is smooth

This is trivial, as $f\vert_{\mathbb{R}_{<0} }\ident 0$

• TODO: The derivative of a constant is 0, the derivative of that is 0, and so forth - link to corresponding pages

For $x>0$, $f$ is smooth

We will show a slightly different result.

• Let $P_m(y)$ be a polynomial of order $m$ in the invariant $y$. $p_m:\mathbb{R}\rightarrow\mathbb{R}$ is the map with $p_m:y\mapsto p(y)$ meaning evaluation at $y\in\mathbb{R}$.

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 $k\eq 1$.

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 $x\in\mathbb{R}_{>0}$ case. We are basically considering $f:\mathbb{R}_{>0}\rightarrow\mathbb{R}$ by $f:x\mapsto e^{-\frac{1}{x} }$

1. Show that $\frac{df}{dx}\eq p_2(\frac{1}{x})e^{-\frac{1}{x} }$ 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 $\dfrac{d^kf}{dx^k}\eq p_{2k}(\tfrac{1}{x})e^{-\frac{1}{x} }$
3. We attempt to show that $\left(\dfrac{d^kf}{dx^k}\eq p_{2k}(\tfrac{1}{x})e^{-\frac{1}{x} }\right)\implies\left(\dfrac{d^{k+1}f}{dx^{k+1} }\eq p_{2(k+1)}(\tfrac{1}{x})e^{-\frac{1}{x} } \right)$
   • $$\frac{d^{k+1}f}{dx^{k+1} }\eq\frac{d}{dx}\left[\frac{d^kf}{dx^k}\right]_x\eq\frac{d}{dx}\left[p_{2k}(\tfrac{1}{x})e^{-\frac{1}{x} }\right]$$ $$\eq e^{-\frac{1}{x} }\left(\frac{d}{dy}\Big[p_{2k}(y)\Big]_{y\eq-\frac{1}{x} }\cdot\frac{d}{dx}\Big[-x^{-1}\Big]\right)+p_{2k}(\tfrac{1}{x})\cdot\frac{d}{dx}\Big[e^{-\frac{1}{x} }\Big]$$
     • Well if we differentiate a polynomial of order $2k$ we get a polynomial of order $2k-1$, so: Switching from $p$ to $P$ 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 $x^{-1}\eq(\tfrac{1}{x})^2$ 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 $2k+2\eq 2(k+1)$ in the subscript of $P$)
   • 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.

Notes

1. Jump up ↑ The details are as follows:
   • $$\frac{df}{dx}$$ is itself a function that takes $x'$ 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 $f$ (wrt $x$) - at $z$ - 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

References