Funzione continua in due variabili

[JackBogno]

Salve, ho un dubbio su questo esercizio:

Sia $f : A ⊆ R × R^n → RR$, con $A$ aperto. Dimostrare che se $f$ è
localmente Lipschitziana rispetto a $x$ in $A$ e $f$ è
continua nella variabile $t$, allora $f$ è continua anche nelle due variabili
$(t, x)$.

Parto dicendo che siccome $f$ è localmente Lipschitziana rispetto ad $x$ allora è continua in $x$, quindi essendo anche continua in $t$, $f$ è continua in $(t, x)$.

Non sono sicuro che sia giusto visto che mi sembra "troppo" semplice, qualcuno può dirmi se ci sono altre considerazioni da fare o se la continuità in $x$ mi fa arrivare subito a conclusione?

Grazie in anticipo.

[javicemarpe]

It is not clear where are the points because you said that $A$ is an open set in $\mathbb{R}\times\mathbb{R}^n$ but then you say that the function is locally Lipschitz with respecto to $X\in A$ and continuous with respect to $t$, so I think you wanted to say that $A$ is an open subset of $\mathbb{R}^n$ or that $A$ is an open subset of $\mathbb{R}\times\mathbb{R}^n$ and $x$ is in $\pi_n (A)$, where $\pi_n$ is the projection in the $n$ last variables.

In any case, if a function is locally Lipschitz in the variable $x$ and continuous in the variable $t$, then is continuous in the variable $(t,x)$. Indeed, let's see that it is continuous in some arbitrary point $(t_0,x_0)\in A$. Let $\varepsilon>0$. As $f$ is locally Lipschitz in $A$ with respect to the variable $x$, we have that there exists an open set $U\subset \pi_n (A)$ (and we can think that $U=B(x_0,\frac{\varepsilon}{2L})$) such that $x_0\in U$ and $$|f(t,x)-f(t,y)|\leq L|x-y|,\quad x,y\in U, t\in \pi_1(A),$$
where $L$ is the Lipschitz constant and $\pi_1$ is the projection in the first variable.

In particular, we have that
$$|f(t,x)-f(t,x_0)|\leq L|x-x_0|<\frac{\varepsilon}{2},\quad x\in U, t\in \pi_1(A).$$

Now, as $f$ is continuous in $t$, there exists a $\delta>0$ associated to $\varepsilon$ such that, if $t\in \pi_1(A)$ is such that $|t-t_0|<\delta$, then
$$|f(t,x)-f(t_0,x)|<\varepsilon/2,\quad x\in \pi_n(A).$$

Now, we have that, if $(t,x)\in (t_0-\delta,t_0+\delta)\times B(x_0,\frac{\varepsilon}{2L})\cap A$, then
\begin{equation*}
\begin{aligned}|f(t,x)-f(t_0,x_0)|&=|f(t,x)-f(t_0,x)+f(t_0,x)-f(t_0,x_0)|\\
&\leq |f(t,x)-f(t_0,x)|+|f(t_0,x)-f(t_0,x_0)|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.
\end{aligned}
\end{equation*}

This means that $f$ is continuous in $(t_0,x_0)$.


Maybe it can be proved more easily using the sequential characterisation of continuity, but now I don't see how to do it.