Conditions for Limit Function to be Limit Minimizing Function of Functional

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Theorem

Let $y$ be a real function.

Let $J\sqbrk y$ be a functional.

Let $\sequence {y_n}$ be a minimizing sequence of $J$.

Let:

$\displaystyle\lim_{n\mathop\to\infty} y_n=\hat y$

Suppose $J$ is lower semicontinuous at $y=\hat y$.


Then:

$\displaystyle J\sqbrk {\hat y}=\lim_{n\mathop\to\infty} J\sqbrk {y_n}$


Proof

By definition of minimizing sequence:

$\displaystyle\inf_y J\sqbrk y=\lim_{n\mathop\to\infty} J\sqbrk {y_n}$

Any mapping from this sequence either minimises the functional or not.

This is true for the limit mapping as well:

$\displaystyle J\sqbrk {\hat y}\ge\inf_y J\sqbrk y$

By assumption, $J$ is lower semicontinuous at $\hat y$:

$\displaystyle\forall\epsilon_1>0:\forall n\in\N:\exists N\in\N:\paren {n>N} \implies\paren {J\sqbrk {\hat y}-J\sqbrk {y_n}<\epsilon_1}$

Hence, for sufficiently large $n$:

$\displaystyle\inf_y J\sqbrk y \le J\sqbrk {\hat y}<J\sqbrk {y_n}+\epsilon_1$

By the definition of minimizing sequence, where the label $n$ has been replaced with $m$:

$\displaystyle\forall\epsilon_2:\forall m\in\N:\exists M\in\N:\paren {m>M} \implies\paren {\size {J\sqbrk {y_m}-\inf_y J\sqbrk y}<\epsilon_2}$

From this, for sufficiently large $m$:

$\displaystyle J\sqbrk {y_m}-\epsilon_2<\inf_y J\sqbrk y$

Then:

$J\sqbrk{y_m}-\epsilon_2<J\sqbrk{\hat y}<J\sqbrk{y_n}+\epsilon_1$

Here $\epsilon_1$, $n$ have similar properties like $\epsilon_2$, $m$, but are otherwise arbitrary and independent.

Let $\epsilon_1=\epsilon_2=\epsilon$, $n=m$, $M=N$. Arbitrariness is still not affected.

Then:

$J\sqbrk{y_n}-\epsilon<J\sqbrk{\hat y}<J\sqbrk{y_n}+\epsilon$

Subtract $J\sqbrk{y_n}$ from all the terms.

This results into:

$-\epsilon<J\sqbrk{\hat y}-J\sqbrk{y_n}<\epsilon$

or

$\size{J\sqbrk{y_n}-J\sqbrk{\hat y} }<\epsilon$

Therefore this relation inherits all the constraints on its values $n$, $N$, $\epsilon$, and by definition is a limit:

$\displaystyle\lim_{n\mathop\to\infty}J\sqbrk{y_n}=J\sqbrk{\hat y}$

$\blacksquare$

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