question about a continuous function and uniform convergence

Question

If $f(x,t)$ is a continuous function on $[a,b]\times[\alpha,\beta]$.$t_0 \in [\alpha,\beta]$.Is $\lim_{t\to t_0}f(x,t)=\varphi(x)$ convergence uniformly about x.Is it true that $\forall \varepsilon>0,\exists \delta$ if $|t-t_0|<\delta$ we have $|f(x,t)-f(x,t_0)|<\varepsilon ,\forall x$

Answer 1

Yes certainly . You know given $ \epsilon >0$ there is a $ \delta > 0 $ so that if $\ |s-s_0|<\delta$ and $ |t-t_0|<\delta $ then $ |f(s,t)-f(s_0,t_0)|< \epsilon $. Now just take x=s=$s_0$ ( since |x-x|=0 <$\delta$ ) to get |f(x,t)-f(x,$t_0$)| < $\epsilon$

Answer 2

Yes, since $f$ is continuous you know that for every $\epsilon > 0$ there exists a $\delta > 0$ such that for all $(s,t) \in [a,b] \times [\alpha, \beta]$ $$\sqrt{(s-x)^2 + (t-t_0)^2} < \delta \implies |f(s,t)-f(x,t_0)| < \epsilon$$ So fix $s$ at $x$ and your result follows.

Answer 3

You may be a little confused about the difference between pointswise convergence ($\delta$ depends on both x and $\epsilon$) and uniform convergence ($\delta$ must hold for all x, so is only in function of $\epsilon$).