Hadamard three-lines Theorem #
In this file we present a proof of a special case of Hadamard's three-lines Theorem.
Main result #
norm_le_interp_of_mem_verticalClosedStrip
: Hadamard three-line theorem onre ⁻¹' [0,1]
: Iff
is a bounded function, continuous onre ⁻¹' [0,1]
and differentiable onre ⁻¹' (0,1)
, then forM(x) := sup ((norm ∘ f) '' (re ⁻¹' {x}))
, that isM(x)
is the supremum of the absolute value off
along the vertical linesre z = x
, we have that∀ z ∈ re ⁻¹' [0,1]
the inequality‖f(z)‖ ≤ M(0) ^ (1 - z.re) * M(1) ^ z.re
holds. This can be seen to be equivalent to the statement thatlog M(re z)
is a convex function on[0,1]
.norm_le_interp_of_mem_verticalClosedStrip'
: Variant of the above lemma in simpler terms. In particular, it makes no mention of the helper functions defined in this file.
Main definitions #
Complex.HadamardThreeLines.verticalStrip
: The vertical strip defined by :re ⁻¹' Ioo a b
Complex.HadamardThreeLines.verticalClosedStrip
: The vertical strip defined by :re ⁻¹' Icc a b
Complex.HadamardThreeLines.sSupNormIm
: The supremum function on vertical lines defined by :sSup {|f(z)| : z.re = x}
Complex.HadamardThreeLines.interpStrip
: The interpolation between thesSupNormIm
on the edges of the vertical strip.Complex.HadamardThreeLines.invInterpStrip
: Inverse of the interpolation between thesSupNormIm
on the edges of the vertical strip.Complex.HadamardThreeLines.F
: Function defined byf
timesinvInterpStrip
. Convenient form for proofs.
Note #
The proof follows from Phragmén-Lindelöf when both frontiers are not everywhere zero.
We then use a limit argument to cover the case when either of the sides are 0
.
The vertical strip in the complex plane containing all z ∈ ℂ
such that z.re ∈ Ioo a b
.
Equations
Instances For
The vertical strip in the complex plane containing all z ∈ ℂ
such that z.re ∈ Icc a b
.
Equations
Instances For
The supremum of the norm of f
on imaginary lines. (Fixed real part)
This is also known as the function M
Equations
- Complex.HadamardThreeLines.sSupNormIm f x = sSup (norm ∘ f '' (Complex.re ⁻¹' {x}))
Instances For
The inverse of the interpolation of sSupNormIm
on the two boundaries.
In other words, this is the inverse of the right side of the target inequality:
|f(z)| ≤ |M(0) ^ (1-z)| * |M(1) ^ z|
.
Shifting this by a positive epsilon allows us to prove the case when either of the boundaries is zero.
Equations
- Complex.HadamardThreeLines.invInterpStrip f z ε = (↑ε + ↑(Complex.HadamardThreeLines.sSupNormIm f 0)) ^ (z - 1) * (↑ε + ↑(Complex.HadamardThreeLines.sSupNormIm f 1)) ^ (-z)
Instances For
A function useful for the proofs steps. We will aim to show that it is bounded by 1.
Equations
- Complex.HadamardThreeLines.F f ε z = Complex.HadamardThreeLines.invInterpStrip f z ε • f z
Instances For
sSup
of norm
is nonneg applied to the image of f
on the vertical line re z = x
sSup
of norm
translated by ε > 0
is positive applied to the image of f
on the
vertical line re z = x
Useful rewrite for the absolute value of invInterpStrip
The function invInterpStrip
is diffContOnCl
.
If f
is bounded on the unit vertical strip, then f
is bounded by sSupNormIm
there.
Alternative version of norm_le_sSupNormIm
with a strict inequality and a positive ε
.
When the function f
is bounded above on a vertical strip, then so is F
.
Proof that F
is bounded by one one the edges.
The interpolation of sSupNormIm
on the two boundaries.
In other words, this is the right side of the target inequality:
|f(z)| ≤ |M(0) ^ (1-z)| * |M(1) ^ z|
.
Note that if (sSupNormIm f 0) = 0 ∨ (sSupNormIm f 1) = 0
then the power is not continuous
since 0 ^ 0 = 1
. Hence the use of ite
.
Equations
- One or more equations did not get rendered due to their size.
Instances For
Rewrite for InterpStrip
when 0 < sSupNormIm f 0
and 0 < sSupNormIm f 1
.
Rewrite for InterpStrip
when 0 = sSupNormIm f 0
or 0 = sSupNormIm f 1
.
Rewrite for InterpStrip
on the open vertical strip.
Hadamard three-line theorem on re ⁻¹' [0,1]
: If f
is a bounded function, continuous on the
closed strip re ⁻¹' [0,1]
and differentiable on open strip re ⁻¹' (0,1)
, then for
M(x) := sup ((norm ∘ f) '' (re ⁻¹' {x}))
we have that for all z
in the closed strip
re ⁻¹' [0,1]
the inequality ‖f(z)‖ ≤ M(0) ^ (1 - z.re) * M(1) ^ z.re
holds.
Hadamard three-line theorem on re ⁻¹' [0,1]
(Variant in simpler terms): Let f
be a
bounded function, continuous on the closed strip re ⁻¹' [0,1]
and differentiable on open strip
re ⁻¹' (0,1)
. If, for all z.re = 0
, ‖f z‖ ≤ a
for some a ∈ ℝ
and, similarly, for all
z.re = 1
, ‖f z‖ ≤ b
for some b ∈ ℝ
then for all z
in the closed strip
re ⁻¹' [0,1]
the inequality ‖f(z)‖ ≤ a ^ (1 - z.re) * b ^ z.re
holds.