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I was wondering how to prove or disprove the following sequence of functions is uniformly convergent

$$ f(n) = frac{nt}{nt+1}, n≥1, t:[0,1] to R$$

So far I have analyzed the limits at $t=0$ and $t=1$ and believe it to be point-wise convergent, but not uniformly convergent. However, I'm not sure how to prove this.

Also, how does the above sequence vary from

$$ f(n) = frac{nt}{n+t}, n≥1, t:[0,1] to R$$
in regards to uniform convergence?

Observe $$f_n(t)=frac{nt}{nt+1}=dfrac{t}{t+dfrac{1}{n}}tobegin{cases}0 , &t=0\1, &tneq0end{cases}$$

For $t$ fixed in $(0,1]$,

$$lim_{nto+infty}f_n(t)=1=f(t)$$

$$lim_{nto+infty}f_n(0)=0=f(0)$$

the pointwise limit $f$ is not continuous at $[0,1]$, the convergence is Not uniform since all $f_n$ are continuous at $[0,1]$.

The sequence

$$f_n(t) = frac{nt}{nt+1}$$

converges (in a point-wise sense) to:

$$f(t) = begin{cases}
0 & text{if}~t=0\
1 & text{if}~0<tleq 1
end{cases}.$$

To check uniform convergence, you need to evaluate $d_n(t) = |f_n(t) - f(t)|$:

$$d_n(t) = begin{cases}
0 & text{if}~t=0\
displaystyle frac{1}{nt+1} & text{if}~0<tleq 1
end{cases}.$$

Notice that:

$$sup_{t in [0,1]} d_n(t) = 1,$$

which does not converge to $0$ as $n$ goes to infinity. Therefore, convergence is not uniform.