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Let $B \subset A$ be Banach spaces with a continuous embedding.

Is the inequality $$ \|b\|_B \leq C \sup_{t > 0} \inf_{\tilde{b} \in B} \{ \|b - \tilde{b}\|_B + t \|\tilde{b}\|_A \} \quad \forall b \in B $$ valid for some $C \geq 1$?

What if the embedding is compact?

For example, take the sequence spaces $B := \ell_2$ and $A := \ell_\infty$ with norms $|\cdot|_\infty \leq |\cdot|_2$. Let $b \in \ell_2$. Let $K \geq 0$ be the $\sup \inf$. Then, for any $t > 0$, the infimizer satisfies $|\tilde{b}|_\infty \leq K / t$, and reduces each component of $b$ by at most $K/t$: $$ \inf_{\tilde{b} \in \ell_2} \{ |b - \tilde{b}|_2 + t |\tilde{b}|_\infty \}^2 \geq \sum_{n} \max\{ 0, |b_n| - K/t\}^2 . $$ This sum goes to $|b|_2^2$ as $t \to \infty$, implying the inequality with $C = 1$.

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Edited. Clearly the reverse inequality is always true with $C = 1$. Here is why: $$\inf_{\tilde{b} \in B} \{ \|b - \tilde{b}\|_B + t \|\tilde{b}\|_A \} \le \|b - 0\|_B + t \|0\|_A = \| b\|_B $$ and hence $$ \sup_{t > 0} \inf_{\tilde{b} \in B} \{ \|b - \tilde{b}\|_B + t \|\tilde{b}\|_A \} \le \|b\|_B $$ for all $b \in B$. Therefore the inequality in your question can only be correct if $C \ge 1$.

Edited again. An incorreect attempt to prove the inequality has been withdrawn.

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  • $\begingroup$ Could you expand on "implying", please? $\endgroup$ – user66081 Feb 8 '15 at 22:05
  • $\begingroup$ Hmm ... you're right with your skepticism, I proved the reverse inequality with $C = 1$. I have edited the response accordingly. $\endgroup$ – Hans Engler Feb 8 '15 at 22:21
  • $\begingroup$ @user66081 - I have entered a complete solution. $\endgroup$ – Hans Engler Feb 9 '15 at 0:54
  • $\begingroup$ @FamousBlueRaincoat - you're right :( $\endgroup$ – Hans Engler Feb 9 '15 at 2:17

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