# Conditions for Limit Function to be Limit Minimizing Function of Functional

## 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:

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

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

Then:

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

## Proof

By definition of minimizing sequence:

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

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

This is true for the limit mapping as well:

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

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

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

Hence, for sufficiently large $n$:

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

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

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

From this, for sufficiently large $m$:

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

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.

Hence:

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

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:

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

$\blacksquare$