Recipe

Compute the L¹, L², and L∞ norms of a function

Use when: you have an explicit, possibly piecewise, formula for f on an interval and need one or more of its three norms

Goal: exact values for \lVert f\rVert_1, \lVert f\rVert_2, and \lVert f\rVert_\infty, computed directly from f's formula

Taught in The L1, L2, and L∞ norms compared

1. 1

   Write f piecewise if a single formula doesn't cover the whole interval, and mark where each piece is positive, negative, or zero.

   Example

   Let f(t)=2 for |t|\le1.5 and f(t)=0 for 1.5<|t|\le\pi. There are two pieces: constant 2 on [-1.5,1.5], and 0 elsewhere.

   Why: \lVert f\rVert_1 needs |f|, and getting each piece's sign right is what "integrating |f|" actually means in practice.

2. 2

   On each piece, integrate |f|, not f itself, then add the pieces together.

   Example

   On [-1.5,1.5], f=2\ge0, so \lVert f\rVert_1=\displaystyle\int_{-1.5}^{1.5}2\,dt=2(3)=6. The zero piece contributes nothing.

   • f never changes sign on a piece — integrate f (or -f) directly, whichever is nonnegative there.
   • f changes sign inside one piece — split that piece at the sign change first, then integrate each half separately.

   Why: \lVert f\rVert_1 is defined as \int|f|, so skipping the absolute value silently cancels area that should add.

3. 3

   Square f first; this removes the sign issue automatically. Integrate over each piece, add the results, then take the square root.

   Example

   \displaystyle\int_{-1.5}^{1.5}2^2\,dt=4(3)=12, so \lVert f\rVert_2=\sqrt{12}=2\sqrt3\approx3.464.

   ⚡ Shortcut

   When f takes only the values 0 and one height h, f^2=h\cdot f on the nonzero piece. That gives \lVert f\rVert_2=\sqrt{h\cdot\lVert f\rVert_1}, with no separate integral needed. Here h=2 and \lVert f\rVert_1=6, so \sqrt{2\cdot6}=2\sqrt3, matching the long way exactly.

   Why: squaring makes every value nonnegative on its own, so the sign bookkeeping Step 2 required disappears completely.

4. 4

   Scan each piece for its largest |f|-value, then take the largest across all pieces; that number is \lVert f\rVert_\infty.

   Example

   The two pieces have magnitudes 2 and 0; the larger is 2, so \lVert f\rVert_\infty=2.

   • f is piecewise-constant — the tallest constant piece's magnitude is the answer directly.
   • f varies continuously within a piece — find that piece's maximum with calculus (critical points and endpoints) before comparing across pieces.

   Why: the supremum is the tightest bound |f(t)| never exceeds, and one missed interior peak invalidates the whole answer.