Is this possible?

$\{f_n(t)\},n\geq1, t\in[0,1],$ is a sequence of absolutely continuous functions with $f_n(0)=f_n(1)=0.$

$$\int_0^1f_n'(t)^2dt\leq C<\infty,$$ but $$\lim_{n\rightarrow\infty}\int_0^1\left(\frac{f_n(t)}{1-t}\right)^2dt=\infty.$$

I strongly believe that this is IMPOSSIBLE, can someone help? Thanks a lot...

  • $\begingroup$ A quick fact: $\|f_n\|^2\leq C$ under the uniform norm. $\endgroup$ – user108871 Sep 17 '14 at 15:03

Let $0<h<1$. We have that:

$$\int_0^h \left(\frac{f_n(t)}{1-t}\right)^2dt= \left[\frac{f_n(t)^2}{1-t}\right]_0^h -2\int_0^h \frac{f_n(t)f_n^{\prime}(t)}{1-t}dt$$ We have: $$ \left[\frac{f_n(t)^2}{1-t} \right]_0^h=\frac{f_n(h)^2}{1-h}=f_n(h)\frac{f_n(h)}{1-h}$$ But $\displaystyle \frac{f_n(h)}{1-h} \to -f_n^{\prime}(1)$ as $h\to 1$, and $f_n(h)\to f_n(1)=0$. Hence we show if $h\to 1$ that $$ \int_0^1 \left(\frac{f_n(t)}{1-t}\right)^2dt= -2\int_0^1 \frac{f_n(t)f_n^{\prime}(t)}{1-t}dt$$

Now we use the Cauchy-Schwarz inequality:

$$\left(\int_0^1 \frac{f_n(t)f_n^{\prime}(t)}{1-t}dt\right)^2\leq \int_0^1 \left(\frac{f_n(t)}{1-t}\right)^2dt\int_0^1 \left(f_n^{\prime}(t)\right)^2dt$$

Hence $\displaystyle \int_0^1 \left(\frac{f_n(t)}{(1-t)}\right)^2 dt=0$ or $$\int_0^1 \left(\frac{f_n(t)}{(1-t)} \right)^2dt\leq 4\int_0^1(f_n^{\prime}(t))^2dt\leq 4c$$ and we are done.

  • $\begingroup$ Way is named Cauchy-Schwartz??? These mathematicians are from different centuries, lol. $\endgroup$ – Masacroso Sep 17 '14 at 17:05
  • $\begingroup$ does holds $\lim_{t\to 1}\frac{ f_n(t)^2}{(1-t)}=0$? to me it is not clear, since $\lim_{t\to 1^-} 1/(1-t)=+\infty$ and $f_n(0)=0$. Your solution does not considerate this. $\endgroup$ – math_man Sep 17 '14 at 17:11
  • $\begingroup$ OK, It is true ; I have edited my answer. $\endgroup$ – Kelenner Sep 17 '14 at 17:30
  • $\begingroup$ Thanks...the solution looks good! I may prefer to show $\lim_{t\rightarrow1}\frac{f_n(t)^2}{1-t}=0.$ This is true since $f_n(t)^2=(\int_t^1f_n'(s)ds)^2\leq (1-t)\int_t^1f_n'(s)^2ds.$ This convergence doesn't have to be uniform in $n.$ The thing is that $\frac{f_n(h)}{1-h}\rightarrow-f_n'(1)$ may not be well-defined at a specific point. $\endgroup$ – user108871 Sep 17 '14 at 17:51
  • $\begingroup$ @math_man: Thank you for the edition. $\endgroup$ – Kelenner Sep 17 '14 at 18:13

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