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to prove a series of function is bounded

ssh

New member
Jun 30, 2012
17
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
Feb 7, 2012
2,702
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.
 

ssh

New member
Jun 30, 2012
17
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
Feb 7, 2012
2,702
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

New member
Jun 30, 2012
17
Thanx now its clear