Schwartz second derivative theorem

Schwartz’s second derivative theorem establishes sufficient conditions for the linearity of a function . Used in the theory of trigonometric series.

Wording

If the function $F(x)$ ${\ displaystyle F (x)}$ continuous in some interval $(a,b)$ ${\ displaystyle (a, b)}$ and $\lim _{h\to 0}{\frac {F(x+h)+F(x-h)-2F(x)}{h^{2}}}=0$ ${\ displaystyle \ lim _ {h \ to 0} {\ frac {F (x + h) + F (xh) -2F (x)} {h ^ {2}}} = 0}$ for all values $x$ ${\ displaystyle x}$ in this interval then $F(x)$ ${\ displaystyle F (x)}$ there is a linear function.

Proof

The expression on the left in the condition of the theorem is called the generalized second derivative of the function $F(x)$ ${\ displaystyle F (x)}$ . If a $F(x)$ ${\ displaystyle F (x)}$ has an ordinary second derivative, then the generalized second derivative is equal to it and there is nothing to prove. Consider the function $\phi (x)=F(x)-F(a)-{\frac {x-a}{b-a}}(F(b)-F(a))$ ${\ displaystyle \ phi (x) = F (x) -F (a) - {\ frac {xa} {ba}} (F (b) -F (a))}$ . Obviously $\phi (a)=0$ ${\ displaystyle \ phi (a) = 0}$ and $\phi (b)=0$ ${\ displaystyle \ phi (b) = 0}$ . To prove the theorem, we show that $\phi (x)=0$ ${\ displaystyle \ phi (x) = 0}$ for all values $x$ ${\ displaystyle x}$ . Let's pretend that $\phi (x)$ ${\ displaystyle \ phi (x)}$ takes positive values. Let be $\phi (c)>0$ ${\ displaystyle \ phi (c)> 0}$ at some point $c$ ${\ displaystyle c}$ . We introduce the function $\psi (x)=\phi (x)-{\frac {1}{2}}\epsilon (x-a)(b-x)$ ${\ displaystyle \ psi (x) = \ phi (x) - {\ frac {1} {2}} \ epsilon (xa) (bx)}$ where $\epsilon$ ${\ displaystyle \ epsilon}$ is a small positive number such that $\psi (x)>0$ ${\ displaystyle \ psi (x)> 0}$ . Function $\psi (x)$ ${\ displaystyle \ psi (x)}$ has a positive upper bound and reaches it, by virtue of its continuity, at some point $x=\xi$ ${\ displaystyle x = \ xi}$ . Obviously $\psi (\xi +h)+\psi (\xi -h)-2\psi (\xi )\leqslant 0$ ${\ displaystyle \ psi (\ xi + h) + \ psi (\ xi -h) -2 \ psi (\ xi) \ leqslant 0}$ . But ${\frac {\psi (\xi +h)+\psi (\xi -h)-2\psi (\xi )}{h^{2}}}={\frac {F(\xi +h)+F(\xi -h)-2F(\xi )}{h^{2}}}+\epsilon$ ${\ displaystyle {\ frac {\ psi (\ xi + h) + \ psi (\ xi -h) -2 \ psi (\ xi)} {h ^ {2}}} = {\ frac {F (\ xi + h) + F (\ xi -h) -2F (\ xi)} {h ^ {2}}} + \ epsilon}$ and with $h\rightarrow 0$ ${\ displaystyle h \ rightarrow 0}$ the right side is committed to $\epsilon$ ${\ displaystyle \ epsilon}$ . A contradiction is obtained. A similar contradiction leads to the assumption that $\phi (x)$ ${\ displaystyle \ phi (x)}$ takes negative values. Consequently, $\phi (x)=0$ ${\ displaystyle \ phi (x) = 0}$ for all values $x$ ${\ displaystyle x}$ and $F(x)$ ${\ displaystyle F (x)}$ there is a linear function.

Literature

E. Titchmarsh Theory of functions, M., Science, 1980.

Schwartz second derivative theorem

Wording

Proof

Literature

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