Example:A smooth function that is not real analytic

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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 we shall prove this by induction starting from $k\eq 1$ .

The proof shall proceed as follows:

Show that $\frac{df}{dx}\eq p_2(\tfrac{1}{x})e^{-\frac{1}{x} }$
Assume that $\frac{d^kf}{dx^k}\eq p_{2k}(\tfrac{1}{x})e^{-\frac{1}{x} }$ holds
Show that (dkfdxk=p2k(1x)e−1x)⟹(dk+1fdxk+1=p2(k+1)(1x)e−1x)
- 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

$f:x\mapsto e^{-\frac{1}{x} }$

Show that dfdx=p2(1x)e−1x 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 $\frac{df}{dx}\eq p_2(\tfrac{1}{x})e^{-\frac{1}{x} }$
Now we assume $\dfrac{d^kf}{dx^k}\eq p_{2k}(\tfrac{1}{x})e^{-\frac{1}{x} }$
We attempt to show that (dkfdxk=p2k(1x)e−1x)⟹(dk+1fdxk+1=p2(k+1)(1x)e−1x)
- dk+1fdxk+1=ddx[dkfdxk]x=ddx[p2k(1x)e−1x]=e−1x(ddy[p2k(y)]y=−1x⋅ddx[−x−1])+p2k(1x)⋅ddx[e−1x]
  - 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
- dk+1fdxk+1=e−1xP2k−1(1x)⋅x−2+P2k(1x)⋅x−2e−1x =e−1x(P2k−1(1x)⋅x−2+P2k(1x)⋅x−2)
  - 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.

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Important work for manifolds, will probably crop up again!

Notes

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.
- dfdx|x=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

[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

[2] $\frac{df}{dx}$ is itself a function that takes $x'$ to $\frac{df}{dx}\Big\vert_{x\eq x'}$

[3] $\frac{df}{dx}\Big\vert_x$ 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.

[4] $\frac{df}{dx}\Big\vert_{x\eq z}$ 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 $\frac{d}{dt}\frac{d}{dx}x^3\Big\vert_{x\eq t}\Big\vert_{t}\eq 6t$

[5] For example $\frac{d}{dx}x^3\Big\vert_{x\eq t}\eq 3t^2$ so $\frac{d}{dt}\frac{d}{dx}x^3\Big\vert_{x\eq t}\Big\vert_{t}\eq 6t$

[6] $\frac{d}{dt}\Big[\text{whatever}\Big]$ to mean $\frac{d}{dt}\text{whatever}\Big\vert_t$

[7] $\frac{d}{dt}\Big[\text{whatever}\Big]_{t}$ to mean $\frac{d}{dt}\text{whatever}\Big\vert_t$ , and

[8] $\frac{d}{dt}\Big[\text{whatever}\Big]_{t\eq p}$ to mean $\frac{d}{dt}\text{whatever}\Big\vert_{t\eq p}$

[9] $\frac{d}{dt}\Big[\text{whatever}\Big]\Big\vert_{t\eq p}$ for example

[Note 1]

Example:A smooth function that is not real analytic

Contents

Example

Proof

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

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

Proof

Notes

References

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Example:A smooth function that is not real analytic

Contents

Example

Proof

For x<0x<0, ff is smooth

For x>0x>0, ff is smooth

Proof

Notes

References

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

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