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]: Iffis 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 offalong 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.reholds. 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 bComplex.HadamardThreeLines.verticalClosedStrip: The vertical strip defined by :re ⁻¹' Icc a bComplex.HadamardThreeLines.sSupNormIm: The supremum function on vertical lines defined by :sSup {|f(z)| : z.re = x}Complex.HadamardThreeLines.interpStrip: The interpolation between thesSupNormImon the edges of the vertical strip.Complex.HadamardThreeLines.invInterpStrip: Inverse of the interpolation between thesSupNormImon the edges of the vertical strip.Complex.HadamardThreeLines.F: Function defined byftimesinvInterpStrip. 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.
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The vertical strip in the complex plane containing all z ∈ ℂ such that z.re ∈ Icc a b.
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The supremum of the norm of f on imaginary lines. (Fixed real part)
This is also known as the function M
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- Complex.HadamardThreeLines.sSupNormIm f x = sSup (norm ∘ f '' (Complex.re ⁻¹' {x}))
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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.
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- Complex.HadamardThreeLines.invInterpStrip f z ε = (↑ε + ↑(Complex.HadamardThreeLines.sSupNormIm f 0)) ^ (z - 1) * (↑ε + ↑(Complex.HadamardThreeLines.sSupNormIm f 1)) ^ (-z)
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A function useful for the proofs steps. We will aim to show that it is bounded by 1.
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- Complex.HadamardThreeLines.F f ε z = Complex.HadamardThreeLines.invInterpStrip f z ε • f z
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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.
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- One or more equations did not get rendered due to their size.
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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.