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Start with the approximate surjectivity with geometric correction argument. For a fixed $y$, put $r_0=y$. Choose $x_j$ successively so thatInduction gives $\Vert r_j\Vert \le k^j\Vert y\Vert$ and $\Vert x_j\Vert \le R{k}^{j-1}\Vert y\Vert$. The series ${\sum }_j{x}_j$ is absolutely convergent in the Banach space $V$, because the sum of its norms is bounded by the convergent geometric series $R\Vert y\Vert /(1-k)$. Write its sum as $x$. The bounded linear operator $T$ commutes with this limit, while $T{\sum }_{j=1}^m{x}_j=y-r_m\to y$. ConsequentlyThe choices need not depend linearly on $y$; no bounded linear right inverse is being asserted.

For the bounded extension, use the restriction bounded linear operator $S:C_b(X;\mathbb{R})\to C_b(Y;\mathbb{R})$ on the Banach spaces of bounded continuous functions, each with its supremum norm. We will construct an approximation to any $g$ with $\Vert g{\Vert }_{\infty }=M>0$. The two setsare disjoint closed sets in $X$, since $Y$ is closed. If both are nonempty, the distance to a set construction defines a continuous functionThe denominator is positive at every point: zero distance to a closed set means membership in that set. Thus $\Vert u{\Vert }_{\infty }\le M/3$, with $u=-M/3$ on $A$ and $u=M/3$ on $B$. On $A$ and $B$, and also on the intervening range $|g|<M/3$, this impliesIf only $A$ is empty take $u=M/3$, if only $B$ is empty take $u=-M/3$, and if both are empty take $u=0$; the same bounds hold. For $M=0$ take $u=0$. This proves the approximation hypothesis with $R=1/3$ and $k=2/3$. Applying the argument above to $S$ yields an element $h$ of the bounded continuous functions with $h{|}_Y=g$ and $\Vert h{\Vert }_{\infty }\le \Vert g{\Vert }_{\infty }$. Restriction gives the reverse inequality, so the extension preserves the exact supremum norm:This proves the bounded metric-space case of the Tietze extension theorem rather than presupposing it.

For an arbitrary real continuous function $f$ on $Y$, let $g=\tanh f$. Its values lie strictly between $-1$ and $1$, although its supremum norm can be $1$. The bounded result supplies an extension $h$ with $|h|\le 1$. Let $D=\{x:|h(x)|=1\}$, a closed set disjoint from $Y$. If $D$ is nonempty and $Y$ is nonempty, putBoth factors are continuous functions; $\eta =1$ on $Y$ and $\eta =0$ on $D$. Off $D$ we have $|h|<1$, and on $D$ we have $q=0$, so $|q(x)|<1$ everywhere. Therefore a real continuous extension isOn $Y$ it equals $f$. If $D$ is empty use $q=h$; if $Y$ is empty simply take $F=0$. This handles unbounded $f$ without requiring a uniform margin between $|g|$ and $1$.