I have a simple question.

Is the Black-Scholes Formula convex with respect to Implied volatility parameter $\sigma$ (for calls or put) ?

When I say Black-Scholes I mean for a call the following one (on Forward price $F_t$):

$$Call (F_t,T-t, K, \sigma^2) = F_t.N(d_1) - K.e^{-r.(T-t)}.N(d_2)$$

$$d_1=\frac{Ln(F_t/K)+1/2.\sigma^2.(T-t)}{\sigma.\sqrt{T-t}}$$ $$d_2=d_1 - \sigma.\sqrt{T-t}$$

and for a put

$$Put (F_t,T-t, K, \sigma^2) = K.e^{-r.(T-t)}.(N(-d_2))-F_t.(N(-d_1)) $$

PS: I know the answer is no but is there a fancy way to prove this (i.e. no brutal force differentiation of the vega)

I have a simple question.

Is the Black-Scholes Formula convex with respect to Implied volatility parameter $\sigma$ (for calls or put) ?

When I say Black-Scholes I mean for a call the following one (on Forward price $F_t$):

$$Call (F_t,T-t, K, \sigma^2) = F_t.N(d_1) - K.e^{-r.(T-t)}.N(d_2)$$

$$d_1=\frac{Ln(F_t/K)+1/2.\sigma^2.(T-t)}{\sigma.\sqrt{T-t}}$$ $$d_2=d_1 - \sigma.\sqrt{T-t}$$

and for a put

$$Put (F_t,T-t, K, \sigma^2) = K.e^{-r.(T-t)}.N(-d_2)-F_t.N(-d_1) $$

PS: I know the answer is no but is there a fancy way to prove this (i.e. no brutal force differentiation of the vega)

I have a simple question.

Is the Black-Scholes Formula convex with respect to Implied volatility parameter $\sigma$ (for calls or put) ?

When I say Black-Scholes I mean for a call the following one (on Forward price $F_t$):

$$Call (F_T, K, \sigma^2) = F_t.N(d_1) - K.e^{-r.(T-t)}.N(d_2)$$

$$d_1=\frac{Ln(F_t/K)+1/2.\sigma^2.(T-t)}{\sigma.\sqrt{T-t}}$$ $$d_2=d_1 - \sigma.\sqrt{T-t}$$

and for a put

$$Put (F_T, K, \sigma^2) = K.e^{-r.(T-t)}.(N(-d_2))-F_t.(N(-d_1)) $$

I have a simple question.

Is the Black-Scholes Formula convex with respect to Implied volatility parameter $\sigma$ (for calls or put) ?

When I say Black-Scholes I mean for a call the following one (on Forward price $F_t$):

$$Call (F_t,T-t, K, \sigma^2) = F_t.N(d_1) - K.e^{-r.(T-t)}.N(d_2)$$

$$d_1=\frac{Ln(F_t/K)+1/2.\sigma^2.(T-t)}{\sigma.\sqrt{T-t}}$$ $$d_2=d_1 - \sigma.\sqrt{T-t}$$

and for a put

$$Put (F_t,T-t, K, \sigma^2) = K.e^{-r.(T-t)}.(N(-d_2))-F_t.(N(-d_1)) $$

PS: I know the answer is no but is there a fancy way to prove this (i.e. no brutal force differentiation of the vega)