# to prove a series of function is bounded

#### ssh

##### New member
Q. If each individual function is bounded and if $$f_n\longrightarrow f$$ uniformly on S, then prove that {fn} is uniformly bounded on S.
Proof : Since each fn is bounded implies $$f_n \leq M_n$$
$$\Longrightarrow f_1\leq M_1, f_2 \leq M_2​,$$ and so on
If M = max {M1, M2,........Mn } then each term is certainly less than M

Please let me know is this approach right?

#### Opalg

##### MHB Oldtimer
Staff member
Q. If each individual function is bounded and if $$f_n\longrightarrow f$$ uniformly on S, then prove that {fn} is uniformly bounded on S.
Proof : Since each fn is bounded implies $$f_n \leq M_n$$
$$\Longrightarrow f_1\leq M_1, f_2 \leq M_2​,$$ and so on
If M = max {M1, M2,........Mn } then each term is certainly less than M

Please let me know is this approach right?
When you write M = max {M1, M2,........Mn }, do you mean that M is the max of the bounds of all the functions $f_1, f_2, f_3, \ldots$, or is it just the max of the bounds of the first n functions?

In the first case, if M is the max of the infinite set $\{M_1,M_2,M_3,\ldots\}$, how do you know that it is finite?

In the second case, if M is the max of the finite set $\{M_1,M_2,M_3,\ldots, M_n\}$ (stopping at $n$), you have only proved something about the first $n$ functions, not the whole sequence.

To give a valid proof of this result, you will need to make use of the fact that the sequence of functions converges uniformly to a limit function.

• Sudharaka

#### ssh

##### New member
I found this answer in a books, but its very confusing, can some one explain this to me clearly.

Ans Since $$f_n \longrightarrow f$$ there exists N such that

$$\arrowvert f_n(x) - f(x) \arrowvert < \epsilon$$ for all n>N and for all x
Mn = 1, 2, ...... be non - negative real numbers such that $$\arrowvert f_n(x)\arrowvert \leq M_n, x\in S$$, n=1,2,.....
Now $$\arrowvert f(x)\arrowvert - \arrowvert f_n(x)\arrowvert \leq \arrowvert f_n(x) - f(x) \arrowvert < 1 x \in S, n<N$$
$$\Longrightarrow \arrowvert f(x) \arrowvert < 1 + \arrowvert f_N0(x) \leq 1 + M_N0 , x \in S$$
this means that f is bounded. it allows follows from the above that for n>N0
$$\arrowvert f_n(x) \arrowvert <1 + \arrowvert f(x) \arrowvert \leq 2 + M_N0$$
let k = max(M1,M2, MN0, 2+MN0)
then $$\arrowvert f_n(x) \arrowvert \leq k, x\in S, n = 1,2,...$$
implies fn is uniformly bounded.

Thanx

#### Opalg

##### MHB Oldtimer
Staff member
I found this answer in a books, but its very confusing, can some one explain this to me clearly.
It is a correct proof. You just have to unpick it carefully and it all makes perfect sense:
Since $$f_n \longrightarrow f$$ there exists N such that

$$\arrowvert f_n(x) - f(x) \arrowvert < \epsilon$$ for all n>N and for all x
More accurately, since $$f_n \longrightarrow f$$ uniformly, given $\varepsilon>0$ there exists $N$ such that

$$| f_n(x) - f(x)| < \varepsilon$$ for all $n> N$ and for all $x$. In particular, if you take $\varepsilon=1$, then there exists $N$ such that

$$\color{red}{ | f_n(x) - f(x)| < 1}$$ for all $\color{red}{n> N}$ and for all $\color{red}x$. Choose some $N_0>N$. Then $\color{blue}{| f_{N_0}(x) - f(x)| < 1 }$ for all $x$.

Mn = 1, 2, ...... be non - negative real numbers such that $$\arrowvert f_n(x)\arrowvert \leq M_n, x\in S$$, n=1,2,.....
Now $$\arrowvert f(x)\arrowvert - \arrowvert f_n(x)\arrowvert \leq \arrowvert f_n(x) - f(x) \arrowvert < 1 x \in S, n<N$$
$$\Longrightarrow \arrowvert f(x) \arrowvert < 1 + \arrowvert f_N0(x) \leq 1 + M_N0 , x \in S$$
this means that f is bounded.
We are told that each $f_n$ is bounded, say $|f_n(x)|\leqslant M_n$ for all $x$. In particular, $\color{blue}{|f_{N_0}(x)|\leqslant M_{N_0}}$ for all $x$. Putting the two blue inequalities together and using the triangle inequality, you see that $|f(x)| = |f_{N_0}(x) + (f(x) - f_{N_0}(x))| \leqslant |f_{N_0}(x)| + |f(x) - f_{N_0}(x)| \leqslant M_{N_0}+1.$

Thus $f$ is bounded, with $M_{N_0}+1$ as a bound.

it allows follows from the above that for n>N0
$$\arrowvert f_n(x) \arrowvert <1 + \arrowvert f(x) \arrowvert \leq 2 + M_N0$$
From the red inequality above, and using the triangle inequality again, you know that $|f_n(x)| = |f(x) + (f_n(x) - f(x))| \leqslant |f(x)| + |f_n(x) - f(x)| \leqslant (M_{N_0}+1) + 1 = M_{N_0}+2$ for all $n>N_0$ and for all $x$.

Thus all the functions from $f_{N_0+1}$ onwards are uniformly bounded, with a bound $M_{N_0}+2$. That just leaves a finite number of functions at the start of the sequence. But each of them is individually bounded, and since there is only a finite number of them, we can just take the maximum of all these bounds to get a bound for the entire sequence:
let $k = \max(M_1,M_2,\ldots, M_{N_0}, 2+M_{N_0})$,
then $|f_n(x)| \leqslant k,\ x\in S,\ n = 1,2,...$
implies $\{f_n\}$ is uniformly bounded.

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• ssh

#### ssh

##### New member
Thanx now its clear