If $f_m \geq 0$ on $[0,T]$, $f_m$ is continuous on $[0,T]$ and $\int_0^T f_m(t)dt \to 0$ as $m \to \infty$ then $f_m(t)\to 0$ as $m \to \infty$.

Question

I was doing this problem in Rudin's book.

And I have a question: Could we proceed the similar problem below by a similar way

If $f_m \geq 0$ on $[0,T]$, $f_m$ is continuous on $[0,T]$ and $\int_0^T f_m(t)dt \to 0$ as $m \to \infty$ then $f_m(t)\to 0$ as $m \to \infty$.

Is this problem true? What do you think?

Thank you so much.

Answer 1

A standard counterexample would be $f_m(t) = t^m$ on the interval $[0,1]$.

Clearly, $$\int_0^1 f_m(t)dt = \frac 1m \to 0 \text{ as } m\to\infty,$$ yet $$\forall m \quad f_m(1)=1.$$

Another, more striking counterexample would be $$f_m(t) = \begin{cases} m-xm^3, &x \in [0,1/m^2],\\0, & x\in [1/m^2,1] \end{cases}$$ These functions are non-negative continuous, their integrals over $[0,1]$ are equal to $\frac {1}{2m}$, yet $f_m(0)\to\infty$ as $m\to\infty$.

In other words, if $f_m$ converges to $0$ in the sense of the Lebesgue space $L^1(0,T)$, we can not guarantee that $f_m$ converges to zero pointwisely or uniformly on $[0,T]$. Fatou's lemma and dominated convergence theorem are classic results for this type of questions.

Answer 2

Answer to your question: No. Just consider $f_m(t) = t^m$ on $[0,1].$

Answer 3

For $n\in\mathbb{N}$ define $f_n : [0,1] \to [0, +\infty\rangle$ as

$$ f(t) = \begin{cases} 1-nt, & \text{if $t \in \left[0,\frac{1}{n}\right]$} \\ 0, & \text{if $t \in \left\langle\frac{1}{n},1\right]$} \end{cases}$$

You can check that

$$\int_{0}^1f_n(t)\,dt = \frac1{2n} \xrightarrow{n\to\infty} 0$$

but $(f_n)_{n=1}^\infty$ does not converge to $0$, not even pointwise, since $f_n(0) = 1, \forall n \in \mathbb{N}$.

Answer 4

In fact, it may the case that $f_m(t)$ does not converge for any $t$.

For simplicity, take $T=1$.

Let $\phi$ be the function created by joining the points $(0,0), ({1 \over 4}, 1), ({3 \over 4}, 1), (1, 0)$.

Pick $n \in \mathbb{N}$ and consider the functions $\phi_{n,k}(t) = \phi(nt-{k \over 4n})$ where $k = -1,0,...,4n(n-1)+1$. Note that for any $n>1$ and any $t \in [0,1]$ there is some $k,k'$ such that $\phi_{n,k}(t) = 1, \phi_{n,k'}(t) = 0$, and $\int_0^1 \phi_{n,k}(t)dt < {1 \over n}$.

Then consider the sequence $\phi_{1,-1}, ..., \phi_{2,-1},, ..., \phi_{3,-1},...$. If we relabel the sequence as $f_1,f_2,...$ then we see that for any $t \in [0,1]$ we have $\liminf_n f_n(t) = 0$ and $\limsup_n f_n(t) = 1$.