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Let $X$ be a Banach space. Let $\phi: [0,\, 1] \to X $ be a sequentially weakly continuous function, that is, $$\forall \,(t_n) \subset [0, 1],\,\,\, t_n \to t \Rightarrow \phi(t_n) \rightharpoonup \phi(t).$$

Consider the function $ f : [0, 1] \to \mathbb{R} $ defined by $f(t) = \|φ(t)\| $ for every $t \in [0,\, 1]$. Prove that

  • $f$ is bounded,

  • there exists the minimum of $f$ on $[0,\, 1]$.

My solution.

Since $\|x\|\geq 0$, the function $f$ is bounded from below and the infimum $\, I:=\inf_{[0,1]} f \,\,$ is finite. Now we apply the direct method in calculus of variations.

Let $t_n$ be a minimizing sequence, that is $f(t_n)\to I$. Since $[0, \,1]$ is compact there exists a subsequence $t_{n_k}$ converging to $\overline{t}\in [0,\,1]$. The norm is weakly lower semicontinuous so $f$ is semicontinuous and in particular $$f(\overline t) \leq \liminf_{k\to\infty}f(t_{n_k})=\lim_{n\to\infty}f(t_{n})=I,$$ but by definition of infimum $I\leq f(\overline t)$ and we get $$f(\overline t)=I.$$

Questions:

  • It is necessary "$X$ is a Banach space" ? I think the answer is no.

  • It is possible to prove that $f$ is upper bounded ? The exercise say "Prove that $f$ is bounded" but I think it's a mistake. I think the author wanted to write "Prove that $f$ is bounded from below" (indeed this is enough for the existence of the minimum).

Thanks in advance.

Ef_Ci
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  • Since you are dealing with metric spaces, $\varphi$ is continuous, hence $f$ too and since $[0, 1]$ is compact, the minimiser exists & $f$ is bounded. – William M. Jun 30 '18 at 21:50
  • The norm is only semicontinuous. Anyway, this is not a problem. I have proved the existence of the minimizer. The main question that remains open is to prove if $f$ it's also upper bounded. – Ef_Ci Jun 30 '18 at 21:53
  • The norm $x \mapsto |x|$ from $X$ into $\Bbb R$ is continuous. – William M. Jun 30 '18 at 21:55
  • Yes, but with respect to the strong convergence and it is only lower semicontinuous with respect to the weak convergence: https://math.stackexchange.com/questions/470217/norm-is-weakly-lower-semicontinuous

    Anyway, I underline that my question is about the boundedness.

     – Ef_Ci Jun 30 '18 at 22:01
  • In that case, you need to state the exercise so that one that is not thinking what you are thinking exactly can understand what you mean. The exercise is ambiguous as stated. – William M. Jun 30 '18 at 22:03
  • The title is "If $\phi:[0,,1]\to X$ be a sequentially weakly continuous, then $f(t)=|\phi(t)|$ is $\textbf{bounded}$". Moreover I explicitly wrote two questions at the end of my post.

    Anyway, you are able to help me ?

     – Ef_Ci Jun 30 '18 at 22:07

2 Answers2

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If this is not a solution, then the exercise as written makes no sense.

Assume, for the sake of reaching an absurd, that $f$ is unbounded. There would exist a sequence $t_n \to t$ such that $f(t_n) > n,$ but since $\varphi(t_n) \to \varphi(t)$ it turns out $\varphi(t_n)$ is bounded (which means $\|\varphi(t_n)\|$ is a bounded sequence of real numbers), hence $f$ is bounded as well, an absurd.

William M.
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  • To be honest, the "weakly" part confused me because "weak topologies" are defined usually on the dual of a space. – William M. Jun 30 '18 at 22:41
  • Thanks. It remains a question: "It is necessary "X is a Banach space" ? I think the answer is no." – Ef_Ci Jun 30 '18 at 22:42
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The answer by Will M. entirely glosses over the essential part. It is true that every weakly bounded sequence is norm bounded, yet this is in no way obvious, but a consequence of the uniform boundedness principle.

To summarize: Since $\phi$ is weakly continuous, its image is weakly compact and thus weakly bounded. By the uniform boundedness principle the image is also norm bounded.

MaoWao
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