Question

Consider the following statements:
(I) The sequence \(\{f_n\}_{n\ge1}\) of functions defined by \(f_n(x)=x^n\) is both point-wise and uniformly convergent on \([0,1]\).
(II) The sequence \(\{f_n\}_{n\ge1}\) of functions defined by \(f_n(x)=\dfrac{\ln(1+nx)}{n}\) is both point-wise and uniformly convergent on \([0,1]\).
Choose the correct answer.

• A. Only (I) is true
• B. Only (II) is true
• C. Both (I) and (II) are true
• D. Neither (I) nor (II) is true

Hint:

Check whether the sup of |f_n - f| over the interval tends to zero, not just the pointwise limit.

Answer 1

The Correct Option is B

Step 1: Consider \(f_n(x)=x^n\) on \([0,1]\). The pointwise limit is \(f(x)=0\) for \(x\in[0,1)\) and \(f(1)=1\), since \(x^n\to0\) for \(x<1\) and stays \(1\) at \(x=1\).

Step 2: Each \(f_n\) is continuous, but the pointwise limit \(f\) is discontinuous at \(x=1\). A uniform limit of continuous functions must itself be continuous, so this convergence cannot be uniform. Indeed \(\sup_{x\in[0,1]}|f_n(x)-f(x)|=\sup_{x\in[0,1)}x^n=1\) for every \(n\), which does not tend to \(0\). So statement (I) is false.

Step 3: Consider \(f_n(x)=\dfrac{\ln(1+nx)}{n}\) on \([0,1]\). At \(x=0\), \(f_n(0)=0\) for all \(n\). For fixed \(x>0\), write \(\ln(1+nx)=\ln n+\ln\left(x+\frac1n\right)\), so \(f_n(x)=\dfrac{\ln n}{n}+\dfrac{\ln(x+1/n)}{n}\to0\) as \(n\to\infty\), since \(\dfrac{\ln n}{n}\to0\). So the pointwise limit is \(f(x)=0\) on all of \([0,1]\).

Step 4: For uniform convergence, \(f_n'(x)=\dfrac{1}{1+nx}>0\), so \(f_n\) is increasing on \([0,1]\) and attains its maximum at \(x=1\): \[\sup_{x\in[0,1]}f_n(x)=f_n(1)=\frac{\ln(1+n)}{n}\] As \(n\to\infty\), \(\ln(1+n)\) grows much slower than \(n\), so this ratio tends to \(0\). Hence \(\sup_x|f_n(x)-0|\to0\), which is exactly uniform convergence. So statement (II) is true.

Step 5: Combining both results, only statement (II) is true.

\[\boxed{\text{Only (II) is true}}\]