Suppose that the sequence $f_j(x)$ on the interval $[0, 1]$ satisfies $|f_j(s) - f_j(t)| \leq |s - t|$ for all $s, t \in [0, 1]$. Further assume that the $f_j$ converge pointwise to a limit function $f$ on the interval $[0, 1]$. Does the series converge uniformly?

This is an exercise from my textbook discussing Weierstrass M-Test.

Attempt: Since $f_j$ satisfies the Lipschitz condition, we know it is uniformly continuous, hence continuous. The continuous mapping of a compact set, $[0, 1]$, is still compact, so the supremum of each $f_j$ is well defined $\forall j$. But I can't really prove that $\displaystyle\sum_{j=1}^{\infty}{M_j} < \infty$, so I don't see how the M-test can be applied.

But if the series really is uniformly convergent, this would imply that $\forall \epsilon > 0$, $\exists N$ such that $\forall p > q > N$ we have $\displaystyle\sum_{k = q+1}^{q}{f_j(x)} < \epsilon$ $\forall x$ which is wierd since this means that $\displaystyle\lim_{j \rightarrow \infty}{f_j(x)} = 0$ for all x, which is a very strong condition that can't be deducted from the fact that $f_j$ being pointwisely convergent to a limit function $f$.

How do I solve this problem? Also, instead of proving the series converges uniformly, can we prove that the sequence converges uniformly?

Any idea would be appreciated.

  • $\begingroup$ I suspect the question intended to ask "does the sequence converge uniformly." Otherwise you could define $f_j(t)=1$ for all $j,t$ and the Lipschitz condition holds but the sum of these does not converge. $\endgroup$
    – Michael
    Dec 10, 2015 at 4:00
  • $\begingroup$ Yes, it is possible to prove the sequence converges uniformly. $\endgroup$
    – Michael
    Dec 10, 2015 at 4:17

1 Answer 1


Choose $\epsilon>0$.

Note that by taking limits that we have $|f(x)-f(y)| \le |x-y|$ for $x,y \in [0,1]$.

Let $G_n = \{ { k \over n } \}_{k=0}^n$. Choose $N$ such that ${1 \over N} < {1 \over 3} \epsilon$. Since $G_N$ is finite, we can find some $N' \ge N$ such that if $n \ge N'$, then $|f(y)-f_n(y)| <{1 \over 3} \epsilon$ for all $y \in G_N$.

For each $x \in [0,1]$, there is some $y \in G_N$ such that $|x-y| \le {1 \over N}$.

Then \begin{eqnarray} |f(x)-f_n(x)| &\le& |f(x)-f(y)| + |f(y)-f_n(y)| + |f_n(y)-f_n(x)| \\ &<& |x-y| + {1 \over 3} \epsilon + |x-y| \\ &\le& {1 \over N} + {1 \over 3} \epsilon + {1 \over N} \\ &<& \epsilon \end{eqnarray}


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