Borderline case of interpolation of Banach spaces Let $B \subset A$ be Banach spaces with a continuous embedding.

Is the inequality
$$
\|b\|_B
\leq
C \sup_{t &gt; 0} \inf_{\tilde{b} \in B} \{ \|b - \tilde{b}\|_B + t \|\tilde{b}\|_A \}
\quad
\forall b \in B
$$
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 $t &gt; 0$, 
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 $C = 1$.
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
$$
\sup_{t &gt; 0} \inf_{\tilde{b} \in B} \{ \|b - \tilde{b}\|_B + t \|\tilde{b}\|_A \}
\le \|b\|_B
$$
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An incorreect attempt to prove the inequality has been withdrawn.
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
$$
\sup_{t &gt; 0} \inf_{\tilde{b} \in B} \{ \|b - \tilde{b}\|_B + t \|\tilde{b}\|_A \}
\le \|b\|_B
$$
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An incorreect attempt to prove the inequality has been withdrawn.