Beedgighref28n

2022-05-03

I have non-negative function $g(y,x)$ that is define using non-negative $f(y,x)$ in the following way:

$\begin{array}{r}g(y,x)={\int}_{0}^{x}f(t,y)dt\end{array}$

I am trying to maximize $\underset{y\in S}{max}g(y,x)$. Using Fatou's lemma we have that

$\begin{array}{r}\underset{y\in S}{max}g(y,x)={\int}_{0}^{x}f(t,y)dt\le {\int}_{0}^{x}\underset{y\in S}{max}f(t,y)dt\end{array}$

I also have that $\underset{y\in S}{max}f(t,y)\le h(t)$ where ${\int}_{0}^{x}h(t)<\mathrm{\infty}$

My question when does the last inequality hold with equally?

What do I have to assume about $f(y,x)$ and $g(x,y)$ for equality to hold?

Can I apply dominated convergence theorem here?

$\begin{array}{r}g(y,x)={\int}_{0}^{x}f(t,y)dt\end{array}$

I am trying to maximize $\underset{y\in S}{max}g(y,x)$. Using Fatou's lemma we have that

$\begin{array}{r}\underset{y\in S}{max}g(y,x)={\int}_{0}^{x}f(t,y)dt\le {\int}_{0}^{x}\underset{y\in S}{max}f(t,y)dt\end{array}$

I also have that $\underset{y\in S}{max}f(t,y)\le h(t)$ where ${\int}_{0}^{x}h(t)<\mathrm{\infty}$

My question when does the last inequality hold with equally?

What do I have to assume about $f(y,x)$ and $g(x,y)$ for equality to hold?

Can I apply dominated convergence theorem here?

Narissiyks

Beginner2022-05-04Added 7 answers

Consider ${y}_{n}$ such that $g({y}_{n},x)\to \underset{y\in S}{max}g(y,x)$.

Now you obtain under the assumption that $\underset{y}{sup}f(t,y)$ is well behaved and bounded

$\underset{y\in S}{max}g(y,x)=\underset{n}{lim}g({y}_{n},x)=\underset{n}{lim}{\int}_{0}^{x}f(t,{y}_{n})\phantom{\rule{thinmathspace}{0ex}}dt={\int}_{0}^{x}\underset{n}{lim}f(t,{y}_{n})\phantom{\rule{thinmathspace}{0ex}}dt$

Therefore the equality

$\begin{array}{r}\underset{y\in S}{max}g(y,x)=\le {\int}_{0}^{x}\underset{y\in S}{max}f(t,y)dt\end{array}$

holds if $\underset{n}{lim}f(t,{y}_{n})=\underset{y\in S}{sup}f(t,y)$.

Under the additional hypothesis that $S$ is compact. Then ${y}_{n}\to {y}^{\ast}$ follows without loss of generality and equality will hold if there is a ${y}^{\ast}$ such that

$f(t,{y}^{\ast})=\underset{y\in S}{max}f(t,y)$

Now you obtain under the assumption that $\underset{y}{sup}f(t,y)$ is well behaved and bounded

$\underset{y\in S}{max}g(y,x)=\underset{n}{lim}g({y}_{n},x)=\underset{n}{lim}{\int}_{0}^{x}f(t,{y}_{n})\phantom{\rule{thinmathspace}{0ex}}dt={\int}_{0}^{x}\underset{n}{lim}f(t,{y}_{n})\phantom{\rule{thinmathspace}{0ex}}dt$

Therefore the equality

$\begin{array}{r}\underset{y\in S}{max}g(y,x)=\le {\int}_{0}^{x}\underset{y\in S}{max}f(t,y)dt\end{array}$

holds if $\underset{n}{lim}f(t,{y}_{n})=\underset{y\in S}{sup}f(t,y)$.

Under the additional hypothesis that $S$ is compact. Then ${y}_{n}\to {y}^{\ast}$ follows without loss of generality and equality will hold if there is a ${y}^{\ast}$ such that

$f(t,{y}^{\ast})=\underset{y\in S}{max}f(t,y)$

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